3x+y
x
=−3
=−y+3
The second equation is solved for xxx, so we can substitute the expression -y+3−y+3minus, y, plus, 3 in for xxx in the first equation:
\begin{aligned} 3\blueD{x}+y &= -3\\\\ 3(\blueD{-y+3})+y&=-3\\\\ -3y+9+y&=-3\\\\ -2y&=-12\\\\ y&=6 \end{aligned}
3x+y
3(−y+3)+y
−3y+9+y
−2y
y
=−3
=−3
=−3
=−12
=6
Plugging this value back into one of our original equations, say x = -y +3x=−y+3x, equals, minus, y, plus, 3, we solve for the other variable:
\begin{aligned} x &= -\blueD{y} +3\\\\ x&=-(\blueD{6})+3\\\\ x&=-3 \end{aligned}
x
x
x
=−y+3
=−(6)+3
=−3
The solution to the system of equations is x=-3x=−3x, equals, minus, 3, y=6y=6y, equals, 6.
We can check our work by plugging these numbers back into the original equations. Let's try 3x+y = -33x+y=−33, x, plus, y, equals, minus, 3.
\begin{aligned} 3x+y &= -3\\\\ 3(-3)+6&\stackrel ?=-3\\\\ -9+6&\stackrel ?=-3\\\\ -3&=-3 \end{aligned}
3x+y
3(−3)+6
−9+6
−3
=−3
=
?
−3
=
?
−3
=−3
Yes, our solution checks out.
Example 2
We're asked to solve this system of equations:
\begin{aligned} 7x+10y &= 36\\\\ -2x+y&=9 \end{aligned}
7x+10y
−2x+y
=36
=9
In order to use the substitution method, we'll need to solve for either xxx or yyy in one of the equations. Let's solve for yyy in the second equation:
\begin{aligned} -2x+y&=9 \\\\ y&=2x+9 \end{aligned}
−2x+y
y
=9
=2x+9
Now we can substitute the expression 2x+92x+92, x, plus, 9 in for yyy in the first equation of our system:
\begin{aligned} 7x+10\blueD{y} &= 36\\\\ 7x+10\blueD{(2x+9)}&=36\\\\ 7x+20x+90&=36\\\\ 27x+90&=36\\\\ 3x+10&=4\\\\ 3x&=-6\\\\ x&=-2 \end{aligned}
7x+10y
7x+10(2x+9)
7x+20x+90
27x+90
3x+10
3x
x
=36
=36
=36
=36
=4
=−6
=−2
Plugging this value back into one of our original equations, say y=2x+9y=2x+9y, equals, 2, x, plus, 9, we solve for the other variable:
\begin{aligned} y&=2\blueD{x}+9\\\\ y&=2\blueD{(-2)}+9\\\\ y&=-4+9 \\\\ y&=5 \end{aligned}
y
y
y
y
=2x+9
=2(−2)+9
=−4+9
=5
The solution to the system of equations is x=-2x=−2x, equals, minus, 2, y=5y=5y, equals, 5.