Solving systems of linear equations by substitution

The solution of a system corresponds to the intersection point of the lines which represent the two linear equations.

Graphic

geogebra plaatje

Substitution method

	Procedure	Example
	When solving a system of two linear equations with two unknowns using the substitution method, we use the following procedure.	Solve the following system: $\left\{\begin{array}{rcl}2 x +4 y+5&=& 0 \\ -3 x +2 y -4&=& 0 \end{array} \right.$
Step 1	In the first equation, express $x$ in $y$ by reduction. In other words, write the first equation in the form $x=\ldots$ .	$\left\{\begin{array}{c}x=-2 y-\frac{5}{2} \\-3 x +2 y -4=0 \end{array} \right.$
Step 2	Substitute the obtained expression for $x$ in the second equation, such that the second equation only contains unknown $y$ .	$\left\{\begin{array}{c}x=-2 y-\frac{5}{2} \\-3 \cdot \left(-2 y - \frac{5}{2}\right) +2 y -4=0 \end{array} \right.$
Step 3	Solve the equation from step 2 for $y$ .	$\begin{array}{rcl}-3 \cdot \left(-2 y - \frac{5}{2}\right) +2 y -4&=&0 \\6 y+\frac{15}{2} +2 y -4&=&0 \\8 y + \frac{7}{2}&=&0 \\8 y &=&-\frac{7}{2}\\y &=&-\frac{7}{16} \end{array}$
Step 4	Determine $x$ using the first equation from step 1 by substituting the value for $y$ found in step 3.	$\begin{array}{rcl} x&=&-2 y-\frac{5}{2} \\ x&=&-2 \cdot- \frac{7}{16}-\frac{5}{2} \\ x&=&-\frac{13}{8}\end{array}$
Step 5	Give the answer in the form $\lineqs{ x & =\;\; \ldots \\ y &=\;\; \ldots }$	$\left\{\begin{array}{rcl}x &=&-\frac{13}{8} \\ y &=&-\frac{7}{16} \end{array}\right.$

Solve the following system of equations.
$\lineqs{x-y&=&-6\cr 6\cdot x+5\cdot y&=&-3\cr }$

$\lineqs{x&=&\displaystyle-3\cr y&=&\displaystyle3\cr }$

Step 1	We reduce the first equation to the form $x=\ldots$ . We find: $\lineqs{x&=&y-6\cr 6\cdot x+5\cdot y&=&-3\cr }$
Step 2	We replace the first equation in the second one. We find: $\lineqs{x&=&y-6\cr 6\cdot\left(y-6\right)+5\cdot y&=&-3\cr }$
Step 3	With help of expanding the brackets, simplification and reduction, we can solve the second equation for unknown $y$ . This is done in the following way: $\begin{array}{rcl}&\lineqs{x&=&y-6\cr 6\cdot\left(y-6\right)+5\cdot y&=&-3\cr }& \\&&\phantom{xxx}\blue{\text{the system we need to solve}} \\ &\lineqs{x&=&y-6\cr 6\cdot y-36+5\cdot y&=&-3\cr }& \\&&\phantom{xxx}\blue{\text{brackets expanded}} \\ &\lineqs{x&=&y-6\cr 11\cdot y-36&=&-3\cr }& \\&&\phantom{xxx}\blue{\text{simplified}} \\ &\lineqs{x&=&y-6\cr 11\cdot y&=&33\cr }& \\&&\phantom{xxx}\blue{\text{both sides of the second equation minus }-36} \\ &\lineqs{x&=&y-6\cr y&=& \displaystyle 3\cr }& \\ &&\phantom{xxx}\blue{\text{both sides of the second equation divided by }11} \end{array}$ Hence, the $y$ -value of the solution is $y=\displaystyle3$ .
Step 4	We now determine $x$ by substituting $y=3$ in the first equation. This is done in the following way: $\begin{array}{rcl}&\lineqs{x&=&\displaystyle 3-6\cr y&=&\displaystyle3\cr }& \\ &&\phantom{xxx}\blue{y=\displaystyle 3 \text{ substituted in the first equation}} \\ &\lineqs{x&=&\displaystyle-3\cr y&=&\displaystyle3\cr}\\ &&\phantom{xxx}\blue{\text{calculated}} \\ \end{array}$

Hence, the solution of the system is: $\lineqs{x&=&\displaystyle-3\cr y&=&\displaystyle3\cr }$