Solving systems of linear equations by substitution

Systems of linear equations: Two equations with two unknowns

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.
\[\lineqs{x-8\cdot y&=&4\cr
8\cdot x-4\cdot y&=&1\cr }\]

#\lineqs{x&=&\displaystyle-{{2}\over{15}}\cr y&=&\displaystyle-{{31}\over{60}}\cr }#

Step 1	We reduce the first equation to the form #x=\ldots#. We find: \[\lineqs{x&=&8\cdot y+4\cr 8\cdot x-4\cdot y&=&1\cr }\]
Step 2	We replace the first equation in the second one. We find: \[\lineqs{x&=&8\cdot y+4\cr 8\cdot\left(8\cdot y+4\right)-4\cdot y&=&1\cr }\]
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: \[\begin{array}{rcl}&\lineqs{x&=&8\cdot y+4\cr 8\cdot\left(8\cdot y+4\right)-4\cdot y&=&1\cr }& \\&&\phantom{xxx}\blue{\text{the system we need to solve}} \\ &\lineqs{x&=&8\cdot y+4\cr 64\cdot y+32-4\cdot y&=&1\cr }& \\&&\phantom{xxx}\blue{\text{brackets expanded}} \\ &\lineqs{x&=&8\cdot y+4\cr 60\cdot y+32&=&1\cr }& \\&&\phantom{xxx}\blue{\text{simplified}} \\ &\lineqs{x&=&8\cdot y+4\cr 60\cdot y&=&-31\cr }& \\&&\phantom{xxx}\blue{\text{both sides of the second equation minus }32} \\ &\lineqs{x&=&8\cdot y+4\cr y&=& \displaystyle -{{31}\over{60}}\cr }& \\ &&\phantom{xxx}\blue{\text{both sides of the second equation divided by }60} \end{array}\] Hence, the #y#-value of the solution is #y=\displaystyle-{{31}\over{60}}#.
Step 4	We now determine #x# by substituting #y=-{{31}\over{60}}# in the first equation. This is done in the following way: \[\begin{array}{rcl}&\lineqs{x&=&\displaystyle 8\cdot -{{31}\over{60}}+4\cr y&=&\displaystyle-{{31}\over{60}}\cr }& \\ &&\phantom{xxx}\blue{y=\displaystyle -{{31}\over{60}} \text{ substituted in the first equation}} \\ &\lineqs{x&=&\displaystyle-{{2}\over{15}}\cr y&=&\displaystyle-{{31}\over{60}}\cr}\\ &&\phantom{xxx}\blue{\text{calculated}} \\ \end{array}\]

Hence, the solution of the system is: \[\lineqs{x&=&\displaystyle-{{2}\over{15}}\cr y&=&\displaystyle-{{31}\over{60}}\cr }\]

New example