The linear-combination method: multiplying one or both equations by constants so that a variable's coefficients become opposites, adding to eliminate it, solving for the survivor, and back-substituting. Includes the case where no multiplication is needed, choosing multipliers and which variable to eliminate, and deciding between this method and substitution.
Subject: Algebra 1 · 65 slides · symbolic lesson
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Title
Algebra 1 · Chapter 7 — Systems of Linear Equations and Inequalities
Solving Linear Systems by Linear Combinations
Objectives
Five outcomes, each one you can test yourself on with a pencil and no answer key.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 7 Systems of Linear Equations and Inequalities — Lesson 7.3 Solving Linear Systems by Linear Combinations §7.3, pp. 402-408 — the lesson these objectives are drawn from
Warm-up
Lesson 7.2 removed a variable by isolating it. This lesson removes one by making it cancel.
Discussion prompt
Look at 4x plus 3y equals 16 and 2x minus 3y equals 8. What happens if you add the left sides together and the right sides together?
Hint: Look at the two y-terms.
Answer:
\[ (4x + 3y) + (2x - 3y) = 16 + 8 \;\Longrightarrow\; 6x = 24 \]
The two y-terms are opposites, so they add to zero and y disappears. What is left is a one-variable equation, which is the same reduction substitution achieved by a different route.
Concept
A linear combination of two equations is obtained by multiplying one or both by a constant if necessary and then adding them. Choosing the constants so that one variable's coefficients are opposites makes that variable vanish.
linear combination — An equation formed by multiplying one or both equations of a system by constants and adding the results, chosen so that one variable is eliminated.
It is useful precisely when no variable is easy to isolate.
Figure (svg): Two equations added so that one variable cancels
McDougal Littell Algebra 1: Concepts and Skills, Ch. 7 Systems of Linear Equations and Inequalities — Lesson 7.3 Solving Linear Systems by Linear Combinations §7.3, pp. 402-402
Section
Section 1
Concept
When a variable has coefficients that are already opposites, adding the two equations makes it disappear and leaves a one-variable equation.
\[ (4x + 3y) + (2x - 3y) = 16 + 8 \;\Longrightarrow\; 6x = 24 \]
Figure (svg): Two equations added so that one variable cancels
McDougal Littell Algebra 1: Concepts and Skills, Ch. 7 Systems of Linear Equations and Inequalities — Lesson 7.3 Solving Linear Systems by Linear Combinations §7.3, pp. 402-402 — Example 1, Add the Equations
Picture it
Plus three y and minus three y.
Figure (svg): Two equations added so that one variable cancels
No multiplication was needed here, so the method reduced to four steps rather than five. Checking for this case first is worth a glance.
Worked example
This is Example 1 from the textbook.
\[ \text{Solve } \; 4x + 3y = 16 \; \text{ and } \; 2x - 3y = 8. \]
Look for opposite coefficients
Why: The y-terms are plus three and minus three.
Add the equations
Why: Six x equals twenty-four.
\[ 6 x = 24 \]
Solve for x
Why: Divide by six.
\[ x = 4 \]
Substitute back
Why: Sixteen plus three y equals sixteen, so y is zero.
\[ (4, 0) \]
Figure (svg): Two equations added so that one variable cancels
\[ (4, 0) \]
Verify: check in both original equations
Why: Sixteen plus zero is sixteen, and eight minus zero is eight. Both hold. Substituting into the second equation to find y would have given the same result, which is a free cross-check when the numbers are convenient.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 7 Systems of Linear Equations and Inequalities — Lesson 7.3 Solving Linear Systems by Linear Combinations §7.3, pp. 402-402
Sorting
Look for a column of opposite coefficients.
Sort into buckets
Sort each system by whether adding the equations directly eliminates a variable.
Four of the six need no preparation at all. Scanning both columns before choosing a method takes a second and often saves the whole multiplication step.
Worked example
The same idea with the other variable.
\[ \text{Solve } \; 5x + 2y = 4 \; \text{ and } \; -5x + 3y = 11. \]
Look for the cancelling column
Why: The x-terms are plus five and minus five.
Add the equations
Why: Five y equals fifteen.
\[ 5 y = 15 \]
Solve for y
Why: Divide by five.
\[ y = 3 \]
Substitute back
Why: Five x plus six is four, so x is negative two fifths.
\[ (-\frac{2}{5}, 3) \]
Figure (svg): The solution to Worked example adding when the x-terms cancel shown as a ladder of expressions, one row per algebraic move
\[ \left(-\tfrac{2}{5}, 3\right) \]
Verify: check in the second original equation
Why: Negative five times negative two fifths is two, plus nine is eleven. The fractional x-coordinate is exactly the kind of answer a graph could not produce, which is why the algebraic methods matter.
Trap
\[ 4x + 3y = 16 \quad \text{and} \quad 2x - 3y = 8 \]
Add the left sides to get 6x and leave the right side as 16
Why: The interesting cancellation is on the left, so attention stays there.
\[ 6x = 16 \quad \text{(wrong)} \]
Adding two equations means adding both sides of each, so the right side becomes twenty-four. Leaving it at sixteen gives x equal to eight thirds, which fails both originals.
\[ 6x = 24 \;\Longrightarrow\; x = 4 \]
Add the left sides together and the right sides together
Why: Each equation says its two sides are equal, so the two sums are equal too.
Writing the addition as a column, with the right-hand sides lined up, makes the omission visible.
Faded example
Left with left, right with right.
Fill in the blanks
(4x + 3y) + (2x - 3y) = 16 + 8 \;\Longrightarrow\; 6x = 24
Why: Adding two equations means adding both of their sides, so the right-hand sides combine to twenty-four. Leaving the right side alone is the standard slip and it produces an x that fails both originals.
Socratic
It looks like a liberty being taken.
Discussion prompt
Explain why adding two equations of a system produces a valid new equation. Then say why the resulting equation's solutions include the system's solution.
Hint: Ask what each equation asserts about the solution pair.
Answer:
At the solution pair, each equation is a true statement: the first left side equals sixteen and the second equals eight. Adding equal quantities to equal quantities preserves equality, which is the addition property from Lesson 3.1 applied to two whole equations rather than to a number.
So any pair satisfying both equations also satisfies their sum, which means the solution survives the operation. The sum has other solutions too — it is one equation in two variables, so its graph is a line — but combined with either original it pins the answer down, which is why one original is used again at the back-substitution step.
Elimination
The system is 2x + y = 7 and 3x - y = 8.
Eliminate the wrong options
What happens when the equations are added?
Survives elimination: A
Why: The y-terms are opposites and cancel, leaving 5x equal to fifteen and so x equal to three. Back-substituting gives y equal to one, so the solution is (3, 1) — and the whole solve took two lines because no multiplication was needed.
Section
Section 2
Concept
When no column has opposite coefficients, multiply one or both equations by constants chosen so that one variable's coefficients become opposites. Then add as before.
Every term of an equation must be multiplied, including the right side.
Figure (svg): Two equations multiplied by different constants to create opposite coefficients
McDougal Littell Algebra 1: Concepts and Skills, Ch. 7 Systems of Linear Equations and Inequalities — Lesson 7.3 Solving Linear Systems by Linear Combinations §7.3, pp. 403-403 — Example 2, Multiply Then Add
Picture it
Four and three, giving twelve.
Figure (svg): Two equations multiplied by different constants to create opposite coefficients
Twelve is the least common multiple of three and four. Using twelve rather than the product of the coefficients keeps every number in the addition smaller.
Worked example
This is Example 2 from the textbook.
\[ \text{Solve } \; 3x + 5y = 6 \; \text{ and } \; -4x + 2y = 5. \]
Choose to eliminate x
Why: The coefficients are three and negative four.
\[ LCM 12 \]
Multiply the equations
Why: The first by four and the second by three.
\[ 12 x\text{ and } -12 x \]
Add them
Why: Twenty-six y equals thirty-nine.
\[ y = 1.5 \]
Substitute back
Why: Negative four x plus three is five, so x is negative a half.
\[ (-0.5, 1.5) \]
Figure (svg): Two equations multiplied by different constants to create opposite coefficients
\[ (-0.5, 1.5) \]
Verify: check in both original equations
Why: For the first: negative one and a half plus seven and a half is six. For the second: two plus three is five. Both hold, and both coordinates are fractional — precisely the case a graph could not have settled.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 7 Systems of Linear Equations and Inequalities — Lesson 7.3 Solving Linear Systems by Linear Combinations §7.3, pp. 403-403
Faded example
Both sides, all terms.
Fill in the blanks
3x + 5y = 6, \; \times 4: \quad 12x + 20y = 24
Why: All three numbers are multiplied by four, including the right-hand side. Multiplying only the left would produce a different line and an answer that fails the original.
Worked example
Sometimes one multiplication is enough.
\[ \text{Solve } \; 2x + 3y = 4 \; \text{ and } \; 4x + 5y = 8. \]
Compare the x-coefficients
Why: Two and four; four is a multiple of two.
Multiply the first by -2
Why: Every term, including the right side.
\[ -4 x - 6 y = -8 \]
Add to the second
Why: Negative y equals zero.
\[ y = 0 \]
Substitute back
Why: Two x equals four, so x is two.
\[ (2, 0) \]
Figure (svg): The solution to Worked example multiplying only one equation shown as a ladder of expressions, one row per algebraic move
\[ (2, 0) \]
Verify: check in both originals
Why: Four plus zero is four, and eight plus zero is eight. Both hold. Because one coefficient divided the other, only a single multiplication was needed — worth checking for before multiplying both equations.
Error analysis
The student multiplied an equation to prepare for adding.
Annotate
On: \( \begin{aligned} 3x + 5y &= 6 \\ \text{multiply by } 4: \quad 12x + 20y &= 6 \end{aligned} \)
Counting the terms before and after multiplying catches this. Every term on both sides changes, or the equation has been replaced by a different one.
Translation
Make one column into opposites.
Match the pairs
Why: When one coefficient divides the other, a single multiplication suffices. Otherwise both equations are multiplied, by the numbers that carry each coefficient up to their least common multiple with opposite signs.
Elimination
Eliminating x from 3x + 5y = 6 and -4x + 2y = 5.
Eliminate the wrong options
Which pair of multipliers makes the x-terms cancel?
Survives elimination: A
Why: Multiplying the first by four gives 12x and the second by three gives negative 12x, which cancel. Option C is the near-miss worth studying: the sizes match and the signs do not, so the column doubles instead of disappearing.
Socratic
The equation looks completely different afterwards.
Discussion prompt
Explain why multiplying every term of an equation by a non-zero constant leaves its solution set unchanged. Then say why the constant must be non-zero.
Hint: Ask what a solution does to the equation.
Answer:
A pair is a solution when substituting it makes the two sides equal. Multiplying both sides by the same number keeps them equal, and dividing by that number afterwards recovers the original — so exactly the same pairs satisfy both versions. This is the multiplication property of equality from Lesson 3.3, applied to a whole equation.
A multiplier of zero would turn the equation into zero equals zero, which every pair satisfies. That destroys all the information the equation carried rather than rewriting it, which is why the property is always stated for non-zero constants — the same exclusion as in Lesson 5.4.
Section
Section 3
Concept
Either variable may be eliminated and the answer is the same. Choosing the one whose coefficients have the smaller least common multiple keeps the numbers manageable.
If one coefficient divides the other, only one equation needs multiplying.
Figure (svg): The same system set up to eliminate either variable
McDougal Littell Algebra 1: Concepts and Skills, Ch. 7 Systems of Linear Equations and Inequalities — Lesson 7.3 Solving Linear Systems by Linear Combinations §7.3, pp. 403-403 — the Solving a Linear System by Linear Combinations summary, step 2
Picture it
Eliminate x, or eliminate y.
Figure (svg): The same system set up to eliminate either variable
Both columns end at the same point. Scanning the four coefficients before choosing is what keeps the numbers small, and small numbers are where the errors are not.
Worked example
Example 2 done the other way round.
\[ \text{Solve } \; 3x + 5y = 6 \; \text{ and } \; -4x + 2y = 5 \; \text{ by eliminating } y. \]
Compare the y-coefficients
Why: Five and two, with least common multiple ten.
\[ LCM 10 \]
Multiply the equations
Why: The first by two and the second by negative five.
\[ 10 y\text{ and } -10 y \]
Add them
Why: Six x plus twenty x is twenty-six x, and twelve minus twenty-five is negative thirteen.
\[ 26 x = -13 \]
Solve and back-substitute
Why: x is negative a half, and y is one and a half.
\[ (-0.5, 1.5) \]
Figure (svg): The same system set up to eliminate either variable
\[ (-0.5, 1.5) \]
Verify: compare the two routes
Why: Eliminating x needed multipliers of four and three; eliminating y needed two and negative five. Both give the same point, and the second route's numbers are slightly smaller — which is the kind of saving the scan is looking for.
Sorting
Compare the coefficients of the variable to eliminate.
Sort into buckets
Sort each situation by how many equations must be multiplied.
Four of the six need one multiplication or none. Checking whether one coefficient divides the other is the second thing to look at, after checking for opposites.
Worked example
Any common multiple works; the smallest is easiest.
\[ \text{Eliminate } x \text{ from } \; 4x + y = 9 \; \text{ and } \; 6x - y = 11 \text{ two ways.} \]
Notice the y-terms already cancel
Why: Plus one and minus one.
Add directly
Why: Ten x equals twenty, so x is two.
\[ x = 2 \]
Consider eliminating x instead
Why: Four and six have least common multiple twelve.
\[ \text{multiply by } 3\text{ and } -2 \]
Compare
Why: The direct addition needed no multiplication at all.
Figure (svg): The solution to Worked example use the least common multiple shown as a ladder of expressions, one row per algebraic move
\[ (2, 1) \]
Verify: check both originals
Why: Eight plus one is nine, and twelve minus one is eleven. Both hold. Scanning found a column already opposite, which made the whole second step unnecessary — the cheapest possible outcome of the scan.
Trap
\[ \text{eliminate } y \text{ from } 5x + 4y = 3 \text{ and } 3x - 6y = 1 \]
Multiply by 6 and 4 to make 24y and -24y
Why: The product of the two coefficients always works, so it is the safe automatic choice.
It does work and it produces numbers four times larger than necessary. The least common multiple of four and six is twelve, so multipliers of three and two would have done the same job.
\[ \times 3 \text{ and } \times 2: \quad 12y \text{ and } -12y \]
Find the least common multiple of the two coefficients
Why: Smaller numbers mean easier addition and fewer places to slip.
The product is never wrong, only heavier, so this is a matter of effort rather than correctness.
Faded example
Least common multiple, opposite signs.
Fill in the blanks
To eliminate y from coefficients 4 and 6, use the least common multiple 12, multiplying by 3 and by -2.
Why: Twelve is the least common multiple of four and six, so multipliers of three and negative two produce 12y and negative 12y. One of the two multipliers must be negative, or the column doubles instead of cancelling.
Elimination
The system is 6x + 5y = 12 and 4x - 2y = 3.
Eliminate the wrong options
Which choice needs the smaller numbers?
Survives elimination: A
Why: Six and four reach twelve with multipliers of two and three, the smallest pair available. The choice is genuinely close here, and the habit of comparing before starting is what matters more than which one wins on any particular system.
Socratic
The signs are as important as the sizes.
Discussion prompt
Explain why the two multipliers cannot both be positive when the original coefficients have the same sign. Then say what happens if both columns are made equal rather than opposite.
Hint: Ask what adding does to two identical terms.
Answer:
Adding is what eliminates, and adding two identical terms doubles them rather than cancelling. So the two coefficients must end up as opposites — same size, different signs — which means one multiplier has to flip a sign if the originals already agree.
If both columns are made equal, adding gives twice the term and nothing is eliminated. Subtracting would then work instead, which is the same as multiplying one equation by negative one first — so the two ways of describing it are the same operation, and only one of them needs to be written down.
Section
Section 4
Concept
When a column has matching rather than opposite coefficients, subtracting one equation from the other eliminates the variable. That is the same as multiplying one equation by negative one and adding.
Writing it as a multiplication by negative one makes the sign changes explicit.
Figure (svg): Two equations with matching coefficients, subtracted rather than added
McDougal Littell Algebra 1: Concepts and Skills, Ch. 7 Systems of Linear Equations and Inequalities — Lesson 7.3 Solving Linear Systems by Linear Combinations §7.3, pp. 403-403 — the summary's requirement that the coefficients be opposites
Picture it
The x-terms are identical, so they subtract away.
Figure (svg): Two equations with matching coefficients, subtracted rather than added
The danger with subtracting is that only the first term gets its sign changed. Multiplying by negative one first and then adding makes every sign change visible.
Worked example
Matching coefficients call for a subtraction.
\[ \text{Solve } \; 3x + 2y = 11 \; \text{ and } \; 3x - 4y = -1. \]
Notice the matching column
Why: Both x-terms are 3x.
Multiply the second by -1 and add
Why: Every term changes sign.
\[ -3 x + 4 y = 1 \]
Add
Why: Six y equals twelve.
\[ y = 2 \]
Back-substitute
Why: Three x plus four is eleven, so x is seven thirds.
\[ (\frac{7}{3}, 2) \]
Figure (svg): Two equations with matching coefficients, subtracted rather than added
\[ \left(\tfrac{7}{3}, 2\right) \]
Verify: check in the second original equation
Why: Three times seven thirds is seven, minus eight is negative one. The check uses the equation that was multiplied, so it also tests whether every sign was flipped correctly.
Faded example
Multiply the whole equation by negative one.
Fill in the blanks
3x - 4y = -1, \; \times (-1): \quad -3x + 4y = 1
Why: All three terms change sign, including the right-hand side. Missing the right side is the commonest version of this error, and it shifts the answer without any visible mistake in the algebra.
Worked example
The last two steps of the summary.
\[ \text{Given } y = 1.5 \text{ for } \; -4x + 2y = 5, \text{ find } x \text{ and check.} \]
Substitute into an original equation
Why: Negative four x plus three equals five.
\[ -4 x + 3 = 5 \]
Solve
Why: Negative four x is two, so x is negative a half.
\[ x = -0.5 \]
Check in the other original
Why: Negative one and a half plus seven and a half is six.
\[ 6 = 6 \]
Report the pair
Why: Both coordinates, x first.
\[ (-0.5, 1.5) \]
Figure (svg): A found value substituted into an original equation to get the other coordinate
\[ (-0.5, 1.5) \]
Verify: notice which equation each step used
Why: The back-substitution used one original and the check used the other, so between them both originals were exercised. Using the same equation twice would leave one of the two untested.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 7 Systems of Linear Equations and Inequalities — Lesson 7.3 Solving Linear Systems by Linear Combinations §7.3, pp. 403-403
Trap
\[ 3x + 2y = 11 \quad \text{minus} \quad 3x - 4y = -1 \]
Write 3x - 3x + 2y - 4y = 11 - 1
Why: The first column's subtraction is done carefully and the rest is copied.
\[ -2y = 10 \quad \text{(wrong)} \]
Subtracting negative 4y adds 4y, and subtracting negative one adds one. Both signs were missed, and the answer for y comes out as negative five rather than two.
\[ -3x + 4y = 1 \text{, then add} \;\Longrightarrow\; 6y = 12 \]
Multiply the whole equation by negative one, then add
Why: Writing the flipped equation out makes every sign change explicit.
Checking in the equation that was flipped is what catches a missed sign, since that is where it happened.
Elimination
Two equations with matching x-coefficients.
Eliminate the wrong options
Subtracting the second equation from the first is the same as which operation?
Survives elimination: A
Why: Subtracting an equation is adding its negative, so multiplying it through by negative one and adding is the same operation written differently. Writing it that way makes every sign change visible, which is why it is worth the extra line.
Prediction
You have found y and need x.
Predict first
Which equation is best for the back-substitution?
Correct: Either original, choosing the one with easier numbers.
\[ -4x + 2(1.5) = 5 \;\Longrightarrow\; x = -0.5 \]
Why: Both originals contain the solution, so either recovers the second coordinate, and picking the one with smaller coefficients saves arithmetic. The combined equation contains only one variable, so it cannot give the other. Using an original also means the multiplied version is never trusted, which keeps a multiplication error from propagating.
Socratic
The multiplied equations are equivalent to them.
Discussion prompt
Explain why the final check uses the original equations rather than the multiplied ones. Then say which of the two originals is the more informative check.
Hint: Ask what could have gone wrong before the check.
Answer:
The multiplied equations are equivalent to the originals only if the multiplication was done correctly. Checking against them would confirm an answer built on a mis-multiplied equation, since both would carry the same error — so the originals are the only independent test available.
The more informative check is the equation not used for the back-substitution, since the answer was constructed to satisfy the other one. Using one original to recover the coordinate and the other to check gets the most out of two substitutions.
Section
Section 5
Concept
Substitution and linear combinations both give exact answers. Substitution suits a system with a coefficient of one; linear combinations suits two equations in standard form with no convenient coefficient.
Either method works on any system; the choice is about effort.
Figure (svg): A guide to choosing between substitution and linear combinations
McDougal Littell Algebra 1: Concepts and Skills, Ch. 7 Systems of Linear Equations and Inequalities — Lesson 7.3 Solving Linear Systems by Linear Combinations §7.3, pp. 402-408 — the lesson's opening remark that isolating a variable is sometimes not easy
Picture it
The coefficients decide.
Figure (svg): A guide to choosing between substitution and linear combinations
Both methods reach the same answers, so neither is more correct. Scanning the four coefficients before starting is what turns a long solve into a short one.
Worked example
The scan takes a few seconds and settles the approach.
\[ \text{Which method for } \; y = 2x + 1, \; 3x + y = 6; \quad x + 4y = 1, \; -2x + 2y = 3; \quad 5x + 3y = 7, \; 4x - 2y = 9? \]
Take the first
Why: One equation is already solved for y.
Take the second
Why: The x in the first equation has coefficient one.
Take the third
Why: No coefficient is one, and both are in standard form.
State the rule
Why: Look for a coefficient of one before anything else.
Figure (svg): A guide to choosing between substitution and linear combinations
\[ \text{sub}, \quad \text{sub}, \quad \text{comb} \]
Verify: try the wrong method on the third system
Why: Isolating x in 5x plus 3y equals 7 gives x equal to seven minus 3y all over five, and every later line carries fifths. Linear combinations multiplies by two and three and works entirely in whole numbers, which is the difference the scan is protecting.
Sorting
Scan the four coefficients.
Sort into buckets
Sort each system by the method that is less work.
Three of each here, and the fourth item is the cheapest of all: its y-terms already cancel, so the whole solve is two lines.
Worked example
The two methods must agree.
\[ \text{Solve } \; 2x + y = 7 \; \text{ and } \; 3x - y = 8 \; \text{ by both methods.} \]
By combinations
Why: The y-terms are opposites, so add: 5x equals 15.
\[ x = 3 \]
Finish
Why: Back-substitute to get y equal to one.
\[ (3, 1) \]
By substitution
Why: Isolate y in the first: y equals negative 2x plus seven.
\[ y = -2 x + 7 \]
Finish
Why: Substituting gives 5x minus seven equals eight, so x is three.
\[ (3, 1) \]
Figure (svg): The solution to Worked example solve one system both ways shown as a ladder of expressions, one row per algebraic move
\[ (3, 1) \]
Verify: compare the two amounts of work
Why: Combinations took two lines because the y-terms already cancelled; substitution took four. Both are correct, and on this system one is plainly quicker — which is exactly what the scan is for.
Trap
\[ 5x + 3y = 7 \quad \text{and} \quad 4x - 2y = 9 \]
Isolate x in the first equation, because substitution is familiar
Why: One method learned well feels safer than choosing between two.
\[ x = \tfrac{7 - 3y}{5} \quad \text{fractions in every later line} \]
Linear combinations handles this in whole numbers. The answer is the same and the route is several times longer, with fifths at every step.
Scan the four coefficients before choosing
Why: A coefficient of one points to substitution; none points to combinations.
Both methods are worth being fluent in precisely so that the choice is available.
Comparison
Fill the blanks from memory before you scroll back.
Comparison matrix
| Substitution (7.2) | Linear combinations (7.3) | |
|---|---|---|
| How a variable is removed | by isolating and replacing | by cancelling in a sum |
| Suits systems with | a coefficient of 1 | no convenient coefficient |
| Risk of fractions | high without a coefficient of 1 | low, since only whole multipliers are used |
Both give exact answers, so neither is more correct. The rows describe effort rather than validity, which is why the choice is worth a few seconds of scanning.
Hypothesis
Predict before you decide.
Predict first
If substitution and linear combinations give different answers for the same system, what has happened?
Correct: An arithmetic error in one of the two routes.
Substituting each candidate answer into both originals identifies which route went wrong.
Why: Both methods use operations that preserve the solution set, so both must reach the point where the two lines cross. Two non-parallel lines meet exactly once, so there is only one answer to find. Working a system both ways is therefore a genuine self-check, and a disagreement locates an error rather than revealing anything about the system.
Socratic
Graphing, substitution and combinations.
Discussion prompt
Say what all three methods of this chapter are doing, described in one sentence that applies to each. Then say what each one does that the others do not.
Hint: Ask what a solution of a system is.
Answer:
All three find the pair satisfying both equations — the point where the two lines cross. Graphing locates it visually, substitution removes a variable by replacement, and combinations removes one by cancellation, but the target is identical in every case.
Graphing is the only one that shows how many solutions exist, which Lesson 7.5 will need. Substitution is the only one that works directly when an equation is given in function form. Combinations is the only one that never requires dividing before the final step, which is why it keeps whole numbers longest. Having all three means every system has a comfortable route.
Comparison
Fill the blanks from memory before you scroll back.
Comparison matrix
| Method | How it works | Best when |
|---|---|---|
| Graphing (7.1) | read the crossing | you want to see how many solutions there are |
| Substitution (7.2) | isolate a variable and replace it | a coefficient is 1 |
| Combinations (7.3) | multiply and add to cancel | no coefficient is 1 |
All three find the same point. Which one to reach for is decided by the shape of the equations, and reading them before starting is the habit worth forming.
Pattern
Whether one equation needs multiplying, both, or neither, the same five moves cover it.
Step two is skipped whenever a column is already opposite, which is worth checking before any multiplying is planned.
OpenStax Elementary Algebra 2e, §5.3 Solve Systems of Equations by Elimination §5.3
Check
Add both sides.
Check your understanding
Adding 4x + 3y = 16 and 2x - 3y = 8 gives which equation?
Answer: A
Why: The y-terms cancel and the x-terms combine to 6x, while the right sides add to twenty-four. Solving gives x equal to four, and back-substituting gives y equal to zero.
Check
Every term, both sides.
Check your understanding
Multiplying 3x + 5y = 6 by 4 gives which equation?
Answer: A
Why: All three numbers are multiplied by four, including the right-hand side. The result is equivalent to the original, so it has exactly the same solutions.
Check
Scan the coefficients.
Check your understanding
For 5x + 3y = 7 and 4x - 2y = 9, which method is less work?
Answer: A
Why: Isolating any variable would introduce fifths, thirds, quarters or halves that persist through the whole solve. Combinations multiplies by whole numbers and adds, keeping every line in integers until the final division.
Real world
A school sells 240 tickets to a concert and takes 1290 dollars. Adult tickets cost 8 dollars and student tickets 3 dollars.
Discussion prompt
Write a system for the number of each kind of ticket, solve it by linear combinations, and interpret the answer. Then say why this method suits the system.
Hint: One equation counts tickets and the other counts dollars.
Answer:
\[ a + s = 240 \qquad 8a + 3s = 1290 \]
\[ \times(-3): \; -3a - 3s = -720 \;\Longrightarrow\; 5a = 570 \;\Longrightarrow\; a = 114 \]
So a hundred and fourteen adult tickets and a hundred and twenty-six student tickets. Checking: the counts add to two hundred and forty, and nine hundred and twelve plus three hundred and seventy-eight is one thousand two hundred and ninety dollars.
Combinations suits it because multiplying the counting equation by negative three makes the s-column cancel in one step, using only whole numbers. Substitution would also work here, since the counting equation has coefficients of one — so this is a system where either method is comfortable, and the scan says so.
Commit first
Answer, then rate your confidence honestly. Confident and wrong is the combination worth finding.
Predict first
To eliminate x from 3x + 5y = 6 and -4x + 2y = 5, what should you multiply by?
Correct: The first by 4 and the second by 3.
\[ 12x + 20y = 24 \quad -12x + 6y = 15 \;\Longrightarrow\; 26y = 39 \]
Why: That gives 12x and negative 12x, which cancel when the equations are added. The first option gives 9x and negative 16x, which are neither equal nor opposite. The third gives 12x and positive 12x, which double instead of cancelling — a near-miss worth noticing, since the sizes are right and the signs are not. Because the original coefficients already have opposite signs, both multipliers should be positive here.
Explain it
They can do substitution and are struggling with a system full of twos and threes.
Discussion prompt
In no more than four sentences, explain what linear combinations does and why it helps here. Then tell them the two things to check while doing it.
Hint: Make a column cancel.
Answer:
A usable answer: instead of getting a letter on its own, you arrange for one letter to disappear when you add the two equations together. Multiply one or both equations so that one column has the same number with opposite signs — then adding kills that column and leaves you one letter to solve for.
Check two things. When you multiply an equation, multiply every term including the number on the right; and when you add, add the right-hand sides too. Both are easy to skip and both change the answer without any visible mistake.
Exit ticket
Name the weakest spot before you close the deck. That is the one worth ten minutes tonight.
Predict first
Which of these would you least want to be handed cold on a quiz tomorrow?
Correct: Whichever you picked is the right answer — and each one has a specific fix.
Why: Multipliers are fixed by finding the least common multiple of the two coefficients and making one multiplier negative if the signs already agree. Multiplying every term is fixed by counting the terms before and after. Subtractions are fixed by writing the equation multiplied by negative one and then adding. The choice of method is fixed by scanning for a coefficient of one before starting. Pick yours and do five of that kind tonight rather than twenty mixed ones.
Connect it up
Do this on paper. It is worth more than rereading the slides.
Draw it
At the top of a page write a system whose coefficients are already opposite in one column, and solve it by adding, in two lines, with the addition written as a column so both sides line up. Underneath, write a system with no convenient coefficients, and solve it by linear combinations: write the two multipliers beside the equations, show both multiplied equations in full, add them as a column, and back-substitute into an original. Check that answer in both originals. To the right, solve the same second system again by eliminating the other variable, and box both answers to show they match, noting which route used smaller numbers. In the lower half, write three systems and beside each write which of the three methods you would use and the reason in five words. Finally, in the margin, write the two things to check when multiplying and adding.
Your two boxed answers must be identical. If they differ, substitute both candidates into the second original equation — the one that fails is the route where a term or a right-hand side went unmultiplied.
Recap
Five things, and the second is where the right-hand side gets forgotten.
| If the question says | Your first move is |
|---|---|
| A column is already opposite | Add the equations directly |
| One coefficient divides the other | Multiply just one equation |
| Neither divides the other | Use the least common multiple |
| The coefficients match rather than oppose | Multiply one equation by -1 |
| No coefficient is 1 | Use linear combinations |
Lesson 7.4 turns the machinery on word problems. With three methods available, the difficulty moves from solving the system to setting it up — deciding what the variables are and what two facts the situation gives you.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 7 Systems of Linear Equations and Inequalities — Lesson 7.3 Solving Linear Systems by Linear Combinations §7.3, pp. 402-408 — everything on these slides traces back here
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