Solving Systems of Equations using Substitution Method
One approach to solving systems of equations is the substitution method. This technique involves choosing one of the equations and expressing one variable in terms of the other. Then, this expression is substituted into the second equation, resulting in a single-variable equation that can be easily solved. Once you determine the value of one variable, you can substitute it back into either of the original equations to find the value of the other variable.
Note: For this lesson, we will just deal with systems of linear equations.
Example 1: Use the substitution method to solve the following system of equations.
$$\eqalign{
2x + y &= – 1 \cr
– 5x + 3y &= 19 \cr} $$
STEP 1: Select the top equation. Solve y in terms of x.
$$2x + y = – 1 \to y = – 2x – 1$$
STEP 2: Write the bottom equation. Replace y with the expressions found in the previous step which is \(y = – 2x – 1\).
$$ – 5x + 3y = 19$$
$$ – 5x + 3\left({\color{red} { – 2x – 1} }\right) = 19$$
STEP 3: Solve the variable x.
$$\eqalign{
– 5x – 6x – 3 &= 19 \cr
– 11x – 3 &= 19 \cr
– 11x &= 19 + 3 \cr
– 11x &= 22 \cr
x &= – 2 \cr} $$
STEP 4: Back substitute the value of x to any of the original equations. Choose the one that is simpler. In this case, we will use the first equation.
$$\eqalign{
2x + y &= – 1 \cr
2\left( \color{red}{ – 2} \right) + y &= – 1 \cr
– 4 + y &= – 1 \cr
y &= 4 – 1 \cr
y &= 3 \cr} $$
STEP 5: Therefore our final solution is
$$\left( { – 2,3} \right)$$
Example 2: Use the substitution method to solve the following system of equations.
$$\eqalign{
– 7x – 3y &= 10 \cr
5x – y &= 18 \cr} $$
In this case, the bottom equation is the simpler one because the absolute value of the coefficient of y is 1.
STEP 1: Select the bottom equation. Solve y in terms of x.
$$\eqalign{
5x – y &= 18 \cr
– y &= – 5x + 18 \cr
y &= 5x – 18 \cr} $$
STEP 2: Write the top equation. Replace y with some expression of x from step 1. Then solve for x.
$$\eqalign{
– 7x – 3y &= 10 \cr
– 7x – 3\left( \color{red}{5x – 18} \right) &= 10 \cr
– 7x – 15x + 54 &= 10 \cr
– 22x &= 10 – 54 \cr
– 22x &= – 44 \cr
x &=2 \cr} $$
STEP 3: Since \(x=2\), we can easily solve for \(y\). Just choose any of the two original equations, plug in the value of \(x\), then solve for \(y\). To prove that it doesn’t matter which one we pick, I will use both of the equations. The value of \(y\) should be the same.
- Using top equation \( – 7x – 3y = 10\)
$$\eqalign{
– 7x – 3y &= 10 \cr
– 7\left( 2 \right) – 3y &= 10 \cr
– 14 – 3y &= 10 \cr
– 3y &= 24 \cr
y &= – 8 \cr} $$
- Using bottom equation \(5x – y = 18\)
$$\eqalign{
5x – y &= 18 \cr
5\left( 2 \right) – y &= 18 \cr
10 – y &= 18 \cr
– y &= 8 \cr
y &= – 8 \cr} $$
Notice that both arrive at \(y=-8\).
STEP 4: Therefore our final answer is
$$\left( { 2,-8} \right)$$
Example 3: Use the substitution method to solve the following system of equations.
$$\eqalign{
2x – 4y &= 8 \cr
5x + 4y &= 13 \cr} $$
If we’re asked to solve this system of equations without being told which method to use, I would definitely choose the elimination method. This is because the coefficients of one of the variables (in this case, \(y\)) have the same absolute value, specifically \(\left| -4 \right| = \left| 4 \right| = 4\).
But we must solve this using the substitution method because that is the direction of this problem.
STEP 1: Select the first equation. Solve x in terms of y.
$$\eqalign{
2x – 4y &= 8 \cr
2x &= 4y + 8 \cr
x &= 2y + 4 \cr} $$
STEP 2: Now write the second equation. Substitute x with the expression containing y from the previous step 1.
$$\eqalign{
5x + 4y &= 13 \cr
5\left( \color{red}{2y + 4} \right) + 4y &= 13 \cr} $$
STEP 3: Solve for the value of y.
$$\eqalign{
10y + 20 + 4y &= 13 \cr
14y + 20 &= 13 \cr
14y &= 13 – 20 \cr
14y &= – 7 \cr
y &= {{ – 7} \over {14}} \cr} $$
$$\boxed{y = {{ – 1} \over 2}}$$
STEP 4: Back substitute the value of y into any of the two original equations then solve for x. I will pick the first equation because it looks less complicated.
$$\eqalign{
2x – 4y &= 8 \cr
2x – 4\left( {{{ – 1} \over 2}} \right) &= 8 \cr
2x + 2 &= 8 \cr
2x &= 8 – 2 \cr
2x &= 6 } $$
$$\boxed{x = 3}$$
STEP 5: Our final solution becomes
$$\left( {3,{{ – 1} \over 2}} \right)$$