### Solving Systems of Equations by Substitution Worksheets

A system of linear equations is composed of two or more equations defined simultaneously. A system of equations in two variables is consists of two equations and two unknowns. The solution to system of equations is a point that satisfies both equations. If graphs are given, the solution set is the point of intersection. If the two graphs intersect at a point then it has one solution. If the lines coincide, then the solution is the set of all points on the line, so the system has infinitely many solutions. If the two lines do not intersect, then there is no real solution. These solutions can be found either by graphing or by algebraic operations. Algebraically, these can be found by elimination, substitution or by comparison.

To solve a system by substitution, follow the following steps:

1. Solve one of the equations $$y$$ in terms of $$x$$or $$x$$ in terms of $$y$$.
2. Replace the resulting equation in step 1 to the second equation.
3. Solve the new equation using the properties of equality, whichever property applies.
4. Once the value of one of the variables is known, use the simplified equation in step 1 to solve for the other variable. You may also use any of the original equations.
5. Write the two values as an ordered pair.
Examples

Solve each system by substitution.
$\small 1. \;\;\displaystyle \left\{\begin{matrix} x+3y=6\\ 2x-y=0 \end{matrix}\right$
Solve $$x+3y=6$$ for $$x$$ in terms of $$y$$.
$$x=6-3y$$

Replace it to the second equation, $$2x-y=0$$.
\small \begin{align*} &2\left ( 6-3y \right )-y=0 \\&12-6y-y=0\\&12-7y=0\\&-7y=12\\&y=\displaystyle \frac{12}{7} \end{align*}
Substitute the value of $$y$$ in $$x=6-3y$$.
\small \begin{align*} &x=6-3\left ( \displaystyle \frac{12}{7} \right )\\&x=6-\displaystyle \frac{36}{7}\\&x=\displaystyle \frac{6}{7} \end{align*}
Therefore, the solution set is $$\left(\displaystyle \frac{6}{7},\displaystyle \frac{12}{7} \right)$$.

$\small 2. \;\;\displaystyle \left\{\begin{matrix} 3x-y=7\\ 4x-y=8 \end{matrix}\right$

Solution:

Solve $$3x-y=7$$ for $$y$$ in terms of $$x$$.
\small \begin{align*} &-y=-3x+7 \\&y=3x-7 \end{align*}

Replace $$y=3x-7$$ to $$4x-2y=8$$.
\small \begin{align*} &4x-2\left ( 3x-7 \right )=8\\&4x-6x+14=8\\&-2x+14=8\\&-2x=8-14\\& -2x=-6\\&\therefore \;\; x=3 \end{align*}
Replace $$x=3$$ to $$y=3x-7$$.
\small \begin{align*} &y=3\left ( 3 \right )-7 \\&y=2 \end{align*}
Therefore, the solution set is $$\left(3,2\right)$$.

Practice Exercises

Solve each system by substitution.
$\small 1. \;\;\displaystyle \left\{\begin{matrix} \displaystyle \frac{1}{2}x=y+3\\ x-y=3 \end{matrix}\right$
$\small 2. \;\;\displaystyle \left\{\begin{matrix} x-3y=8\\ 2x+3y=12 \end{matrix}\right$
$\small 3. \;\;\displaystyle \left\{\begin{matrix} 2x-5y=10\\ x+y=5 \end{matrix}\right$
$\small 4. \;\;\displaystyle \left\{\begin{matrix} x-5y=10\\ x+y=2 \end{matrix}\right$