Substitution Method Practice

We are given that ‍$y = {x+4}$. Let's substitute this expression into the first equation and solve for ‍‍$x$ as follows:

‍$\begin{aligned} 12x-5{y}&=-20\\\\12x-5\cdot({x+4})&=-20\\\\12x-5x-20& = -20\\\\7x&=0\\\\x&=0\end{aligned}$

Since we now know that ‍$x=0$, we can substitute this value into the second equation to solve for ‍$y$ as follows:

$\begin{aligned} y &=  {x}+4 \\\\y&={0}+4\\\\y&=4 \end{aligned}$

This is the solution of the system:

$\begin{aligned}&x = 0\\\\&y=4\end{aligned}$

We are given that ‍$x = {y+6}$. Let's substitute this expression into the first equation and solve for ‍‍$y$ as follows:

‍$\begin{aligned} 13{x}-6y &= 22\\\\13\cdot({y+6})-6y&=22\\\\13y+78-6y&=22\\\\7y&=-56\\\\y&=-8\end{aligned}$

Since we now know that ‍$y=-8$, we can substitute this value into the second equation to solve for ‍$x$ as follows:

$\begin{aligned} x &= {y}+6 \\\\x&={-8}+6\\\\x&=-2 \end{aligned}$

This is the solution of the system:

$\begin{aligned}&x = -2\\\\&y= -8\end{aligned}$

We are given that ‍$y = {3x+15}$. Let's substitute this expression into the first equation and solve for ‍‍$x$ as follows:

‍$\begin{aligned} -4x+7{y}&=20\\\\-4x+7\cdot({3x+15})&=20\\\\-4x+21x+105& = 20\\\\17x&=-85\\\\x&=-5\end{aligned}$

Since we now know that ‍$x=-5$, we can substitute this value into the second equation to solve for ‍$y$ as follows:

$\begin{aligned} y &= 3\cdot {x}+15 \\\\y&=3\cdot({-5})+15\\\\y&=0 \end{aligned}$

This is the solution of the system:

$\begin{aligned}&x = -5\\\\&y=0\end{aligned}$

We are given that ‍$x = {y+3}$. Let's substitute this expression into the first equation and solve for ‍‍$y$ as follows:

‍$\begin{aligned} 6{x}-5y &= 15\\\\6\cdot({y+3})-5y&=15\\\\6y+18-5y&=15\\\\y&=-3\\\\\end{aligned}$

Since we now know that ‍$y=-3$, we can substitute this value into the second equation to solve for ‍$x$ as follows:

$\begin{aligned} x &= {y}+3 \\\\x&={-3}+3\\\\x&=0 \end{aligned}$

This is the solution of the system:

$\begin{aligned}&x = 0\\\\&y= -3\end{aligned}$