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
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.# |
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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} |
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.# |
#\lineqs{x&=&2\cr y&=&-2\cr }#
Step 1 | We reduce the first equation to the form #x=\ldots#. We find: \[\lineqs{x&=&2\cdot y+6\cr -6\cdot x-3\cdot y&=&-6\cr }\] |
Step 2 | We substitute the first equation in the second one. We find: \[\lineqs{x&=&2\cdot y+6\cr -6\cdot\left(2\cdot y+6\right)-3\cdot y&=&-6\cr }\] |
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: \[\begin{array}{rcl}&\lineqs{x&=&2\cdot y+6\cr -6\cdot\left(2\cdot y+6\right)-3\cdot y&=&-6\cr }& \\&&\phantom{xxx}\blue{\text{the system we need to solve}} \\ &\lineqs{x&=&2\cdot y+6\cr -12\cdot y-36-3\cdot y&=&-6\cr }& \\&&\phantom{xxx}\blue{\text{brackets expanded}} \\ &\lineqs{x&=&2\cdot y+6\cr -15\cdot y-36&=&-6\cr }& \\&&\phantom{xxx}\blue{\text{simplified}} \\ &\lineqs{x&=&2\cdot y+6\cr -15\cdot y&=&30\cr }& \\&&\phantom{xxx}\blue{\text{both sides of the second equation minus }-36} \\ &\lineqs{x&=&2\cdot y+6\cr y&=&-2\cr }& \\ &&\phantom{xxx}\blue{\text{both sides of the second equation divided by }-15} \end{array}\] Hence, the #y#-value of the solution is #y=-2#. |
Step 4 | We now determine #x# by substituting #y=-2# in the first equation. This is done in the following way: \[\begin{array}{rcl}&\lineqs{x&=&2\cdot -2+6\cr y&=&-2\cr }& \\ &&\phantom{xxx}\blue{y=-2 \text{ substituted in the first equation}} \\ &\lineqs{x&=&2\cr y&=&-2\cr}\\ &&\phantom{xxx}\blue{\text{calculated}} \\ \end{array}\] |
Hence, the solution of the system is: \[\lineqs{x&=&2\cr y&=&-2\cr }\]
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