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# Solving Systems of Linear Equations using Substitution

## Solving Systems of Linear Equations using Substitution 1.1 - Solution

a

We want to solve the following system of linear equations by substitution. $\begin{cases}x=10-3y & \, \text {(I)}\\ y=x-2 & \text {(II)}\end{cases}$ When solving a system of equations using substitution, there are three steps.

1. Isolate a variable in one of the equations.
2. Substitute the expression for that variable into the other equation and solve.
3. Substitute this solution into one of the equations and solve for the value of the other variable.
For this exercise, $x$ is already isolated in Equation I. $x=10-3y$ Let's substitute this into Equation II, $y=x-2,$ and solve the resulting equation for $y.$
$y=x-2$
$y={\color{#0000FF}{10-3y}}-2$
$y=8-3y$
$4y=8$
$y=2$
To find the value of $x,$ we need to substitute $y=2$ into either one of the equations in the given system. Let's use the first equation.
$x=10-3y$
$x=10-3\cdot {\color{#0000FF}{2}}$
$x=10-6$
$x=4$
The solution, or point of intersection, to this system of equations is the point $(4,2).$
b
Here we want to solve the following system of linear equations. $\begin{cases}2x-2y=12 & \, \text {(I)}\\ x-4=1 & \text {(II)}\end{cases}$ Let's start by isolating $x$ in the equation II. $x-4=1 \quad \Leftrightarrow \quad x=5$ Now that we've isolated $x,$ we can substitute its value into $2x-2y=12$ and solve the resulting equation for $y.$
$2x-2y=12$
$2\cdot{\color{#0000FF}{5}}-2y=12$
$10-2y=12$
$\text{-} 2y=2$
$y=\text{-} 1$
The solution, or point of intersection, to this system of equations is the point $(5,\text{-}1).$