Solving a System of Equations by Substitution

Solve a system of simultaneous equations by substitution by first expressing one variable in terms of the other. Using these equations as an example:

x - y = 5 \\ 3x + 2y = 5

Re-arrange the simplest equation to work with and use this to insert into the second. In this case, adding ​y​ to both sides of the first equation gives:

x = y + 5

Use the expression for ​x​ in the second equation to produce an equation with a single variable. In the example, this makes the second equation:

3 × (y + 5) + 2y = 5 \\ 3y + 15 + 2y = 5

Collect the like terms to get:

5y + 15 = 5

Re-arrange and solve for ​y​, starting by subtracting 15 from both sides:

5y = 5 - 15 = -10

Dividing both sides by 5 gives:

y = \frac{-10}{5} = -2

So ​y​ = −2.

Insert this result into either equation to solve for the remaining variable. At the end of step 1, you found that:

x = y + 5

Use the value you found for ​y​ to get:

x = -2 + 5 = 3

So ​x​ = 3 and ​y​ = −2.

Tips

• ​Check Your Answers​

  It’s good practice to ​always​ check that your answers make sense and work with the original equations. In this example, ​x​ – ​y​ = 5, and the result gives 3 – (−2) = 5, or 3 + 2 = 5, which is correct. The second equation states: 3​x​ + 2​y​ = 5, and the result gives 3 × 3 + 2 × (−2) = 9 – 4 = 5, which is again correct. If something doesn’t match up at this stage, you have made a mistake in your algebra.

Solving a System of Equations by Elimination

Look at your equations to find a variable to remove:

x - y = 5 \\ 3x + 2y = 5

In the example, you can see that one equation has -​y​ and the other has +2​y​. If you add twice the first equation to the second one, the ​y​ terms would cancel out and ​y​ would be eliminated. In other cases (e.g., if you wanted to eliminate ​x​), you can also subtract a multiple of one equation from the other.

Multiply the first equation by two to prepare it for the elimination method:

2 × (x - y) = 2 × 5

So

2x - 2y = 10

Eliminate your chosen variable by adding or subtracting one equation from the other. In the example, add the new version of the first equation to the second equation to get:

3x + 2y + (2x - 2y) = 5 + 10 \\ 3x + 2x + 2y - 2y = 15

So this means:

5x = 15

Solve for the remaining variable. In the example, divide both sides by 5 to get:

x = \frac{15}{5} = 3

As before.

Like in the previous approach, when you have one variable, you can insert this into either expression and re-arrange to find the second. Using the second equation:

3x + 2y = 5

So, since ​x​ = 3:

3 × 3 + 2y = 5 \\ 9 + 2y = 5

Subtract 9 from both sides to get:

2y = 5 - 9 = -4

Finally, divide by two to get:

y = \frac{-4}{2} = -2

Solving a System of Equations by Graphing

Solve systems of equations with minimal algebra by graphing each equation and looking for the ​x​ and ​y​ value where the lines intersect. Convert each equation to slope-intercept form (​y​ = ​mx​ + ​b​) first.

The first example equation is:

x - y = 5

This can be converted easily. Add ​y​ to both sides and then subtract 5 from both sides to get:

y = x - 5

Which has a slope of ​m​ = 1 and a ​y​-intercept of ​b​ = −5.

The second equation is:

3x + 2y = 5

Subtract 3​x​ from both sides to get:

2y = -3x + 5

Then divide by 2 to get the slope-intercept form:

y = \frac{-3x}{2} + \frac{5}{2}

So this has a slope of ​m​ = -3/2 and a ​y​-intercept of ​b​ = 5/2.

Use the ​y​ intercept values and the slopes to plot both lines on a graph. The first equation crosses the ​y​ axis at ​y​ = −5, and the ​y​ value increases by 1 every time the ​x​ value increases by 1. This makes the line easy to draw.

The second equation crosses the ​y​ axis at 5/2 = 2.5. It slopes downwards, and the ​y​ value decreases by 1.5 every time the ​x​ value increases by 1. You can calculate the ​y​ value for any point on the ​x​ axis using the equation if it’s easier.

Locate the point where the lines intersect. This gives you both the ​x​ and ​y​ coordinates of the solution to the system of equations.