M8MATH-5.3

Solving Simple Systems by Substitution

Learn how to solve systems of linear equations using substitution when one equation is already solved for y, or when you can easily solve for y first.

What you'll do in this lesson

A voice-first session with the Crimsora tutor on Solving Simple Systems by Substitution, then targeted practice and FRQs — with the tutor adapting to where you get stuck.

What this lesson covers

When two linear equations describe the same situation, their solution is the point where they intersect. You already know how to find that point by graphing, but graphing only gives you an approximate answer. In this lesson, you'll learn substitution: a method that finds the exact solution algebraically. Substitution works especially well when at least one equation is already solved for y, or when you can solve for y quickly.

What Substitution Means

Substitution means replacing one variable with an expression involving the other variable. In a system where both equations equal y, you can set them equal to each other. For example, if y=2x+1y = 2x + 1 and y=x+7y = -x + 7, then 2x+1=x+72x + 1 = -x + 7. You've eliminated y and created a single equation in x that you can solve. Once you find x, you substitute that number back into either original equation to find y. This is called back-substitution. The key insight is that y represents the same value in both equations, so whatever makes y equal on one side must also work on the other.

Method When Both Equations Equal y

When both equations are already written as y=(expression)y = (\text{expression}), the substitution step is straightforward. Set the two expressions equal: if y=2x+3y = 2x + 3 and y=x+9y = -x + 9, write 2x+3=x+92x + 3 = -x + 9. Solve for x: add x to both sides to get 3x+3=93x + 3 = 9, then subtract 3 to get 3x=63x = 6, so x=2x = 2. Now back-substitute x into either equation. Using the first: y=2(2)+3=7y = 2(2) + 3 = 7. Check in the second: y=(2)+9=7y = -(2) + 9 = 7. Both give the same y, confirming your solution is (2,7)(2, 7). Always substitute back into at least one equation to verify; this catches arithmetic errors before you finish.

Method When Only One Equation Equals y

Sometimes one equation already equals y, but the other doesn't. For example, y=3xy = 3x and 2x+y=102x + y = 10. Substitute the expression 3x3x for y in the second equation: 2x+3x=102x + 3x = 10, which simplifies to 5x=105x = 10, giving x=2x = 2. Then back-substitute into the first equation: y=3(2)=6y = 3(2) = 6. So the solution is (2,6)(2, 6). The method is the same: use the solved-for equation to replace the variable in the other equation, solve, and then find the other coordinate. This avoids having to solve the second equation for y yourself, saving a step.

Solving for y First (When Needed)

If neither equation is solved for y, you can solve one of them for y before substituting. For example, in the system x+y=5x + y = 5 and 2xy=12x - y = 1, solve the first for y: y=5xy = 5 - x. Then substitute into the second: 2x(5x)=12x - (5 - x) = 1. Distribute carefully: 2x5+x=12x - 5 + x = 1, so 3x=63x = 6 and x=2x = 2. Back-substitute: y=52=3y = 5 - 2 = 3. The solution is (2,3)(2, 3). Choose the equation that is easiest to solve for y—often the one where y has a coefficient of 1 or −1. This strategy saves you from messy fractions later.

Why Substitution Gives Exact Answers

Graphing a system finds the intersection point visually, but your accuracy depends on how carefully you plot and read the graph. Substitution is algebraic, so it produces exact coordinates expressed as numbers or simple fractions. For the system y=12x+1y = \frac{1}{2}x + 1 and y=2x+6y = -2x + 6, graphing might place the solution near (2,2)(2, 2), but substitution gives the exact point. Set 12x+1=2x+6\frac{1}{2}x + 1 = -2x + 6: multiply by 2 to get x+2=4x+12x + 2 = -4x + 12, so 5x=105x = 10 and x=2x = 2. Then y=12(2)+1=2y = \frac{1}{2}(2) + 1 = 2. The exact solution is (2,2)(2, 2), no approximation needed.

Key terms

Substitution.
A method for solving a system by replacing one variable with an expression equal to it, reducing the system to one equation in one unknown.
Back-substitution.
The step of replacing a variable in one of the original equations with the value you found, in order to solve for the other variable.
System of equations.
Two or more equations with the same variables, where a solution must satisfy all equations at the same time.
Solution of a system.
An ordered pair (x, y) that makes both equations true when substituted in.
Equivalent equations.
Equations that have the same solution; you can transform one into the other without changing the solution set.
Eliminate a variable.
To remove one variable from an equation, usually by substitution or by combining equations so one variable cancels.

Worked example

Solve the system: y=2x3y = 2x - 3 and y=x+6y = -x + 6.
Step 1: Recognize that both equations equal y. This is the perfect setup for substitution.

Step 2: Set the two expressions for y equal to each other:2x3=x+62x - 3 = -x + 6Step 3: Solve for x. Add x to both sides:3x3=63x - 3 = 6Step 4: Add 3 to both sides:3x=93x = 9x=3x = 3Step 5: Back-substitute x=3x = 3 into one of the original equations. Use the first:y=2(3)3=63=3y = 2(3) - 3 = 6 - 3 = 3Step 6: Check by substituting into the second equation:y=(3)+6=3y = -(3) + 6 = 3

The solution is (3,3)(3, 3). Both equations give y=3y = 3 when x=3x = 3, so the point lies on both lines.

Practice questions

Solve the system y=x+5y = -x + 5 and y=2x1y = 2x - 1 using substitution.

Answer: (2,3)(2, 3)

Set the expressions equal: x+5=2x1-x + 5 = 2x - 1. Add x to both sides: 5=3x15 = 3x - 1. Add 1: 6=3x6 = 3x, so x=2x = 2. Back-substitute: y=(2)+5=3y = -(2) + 5 = 3. Check: y=2(2)1=3y = 2(2) - 1 = 3. ✓ The solution is (2,3)(2, 3).
Solve the system y=4xy = 4x and x+y=10x + y = 10 using substitution. Show all your work.

Answer: (2,8)(2, 8)

The first equation already equals y, so substitute 4x4x for y in the second equation: x+4x=10x + 4x = 10. Combine: 5x=105x = 10, so x=2x = 2. Back-substitute into the first equation: y=4(2)=8y = 4(2) = 8. Verify in the second: 2+8=102 + 8 = 10. ✓ The solution is (2,8)(2, 8).
Which equation should you solve for y first in the system 2x+y=82x + y = 8 and x2y=1x - 2y = 1? Why?

Answer: Solve 2x+y=82x + y = 8 for y. Subtracting 2x gives y=82xy = 8 - 2x.

The first equation has y with a coefficient of 1, so solving for y requires only one step: y=82xy = 8 - 2x. The second equation has y with a coefficient of −2, which would require dividing by −2 and create a fraction. Choosing the easier equation saves work and reduces the chance of error.

FAQ

What if I substitute into the wrong equation?
You will still get the correct answer. The order does not matter because both original equations describe the same relationship. However, it is often easier to substitute into the equation that is already solved for the variable, since that equation is simpler.
Do I have to use the same equation for back-substitution as I used for the first substitution?
No. You can back-substitute into either original equation—you will get the same value for the variable. It is often wise to back-substitute into the equation you did not use for the first step, to double-check your work.
When should I use substitution instead of graphing?
Substitution is better when you need an exact answer or when the solution involves fractions or decimals. Graphing is faster for getting a rough picture of the solution, but substitution always gives the precise coordinates.
Why is substitution called 'substitution'?
Because you substitute (replace) one variable with an expression, reducing the number of variables in play. Once you have one equation with one unknown, it becomes much easier to solve.

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