yes
step-by-step explanation:
we're asked to solve this system of equations:
\begin{aligned} 2y+7x & = -5 5y-7x & = 12 \end{aligned}
2y+7x
5yβ7x
β
=β5
=12
β
we notice that the first equation has a 7x7x7, x term and the second equation has a -7xβ7xminus, 7, x term. these terms will cancel if we add the equations togetherβthat is, we'll eliminate the xxx terms:
2y+7x+ 5yβ7x7y+0=β5=12=7
solving for yyy, we get:
\begin{aligned} 7y+0 & =7 7y & =7 y & =\goldd{1} \end{aligned}
7y+0
7y
y
β
=7
=7
=1
β
plugging this value back into our first equation, we solve for the other variable:
\begin{aligned} 2y+7x & = -5 2\cdot \goldd{1}+7x & = -5 2+7x& =-5 7x& =-7 x& =\blued{-1} \end{aligned}
2y+7x
2β
1+7x
2+7x
7x
x
β
=β5
=β5
=β5
=β7
=β1
β
the solution to the system is x=\blued{-1}x=β1x, equals, start color blued, minus, 1, end color blued, y=\goldd{1}y=1y, equals, start color goldd, 1, end color goldd.
we can check our solution by plugging these values back into the the original equations. let's try the second equation:
\begin{aligned} 5y-7x & = 12 5\cdot\goldd{1}-7(\blued{-1}) & \stackrel ? = 12 5+7 & = 12 \end{aligned}
5yβ7x
5β
1β7(β1)
5+7
β
=12
=
?
12
=12
β
yes, the solution checks out.
if you feel uncertain why this process works, check out this intro video for an in-depth walkthrough.
example 2
we're asked to solve this system of equations:
\begin{aligned} -9y+4x - 20& =0 -7y+16x-80& =0 \end{aligned}
β9y+4xβ20
β7y+16xβ80
β
=0
=0
β
we can multiply the first equation by -4β4minus, 4 to get an equivalent equation that has a \purpled{-16x}β16xstart color purpled, minus, 16, x, end color purpled term. our new (but equivalent! ) system of equations looks like this:
\begin{aligned} 36y\purpled{-16x}+80& =0 -7y+16x-80& =0 \end{aligned}
36yβ16x+80
β7y+16xβ80
β
=0
=0
β
adding the equations to eliminate the xxx terms, we get:
36yβ16x+80+ β7y+16xβ8029y+0β0=0=0=0
solving for yyy, we get:
\begin{aligned} 29y+0 -0& =0 29y& =0 y& =\goldd 0 \end{aligned}
29y+0β0
29y
y
β
=0
=0
=0
β
plugging this value back into our first equation, we solve for the other variable:
\begin{aligned} 36y-16x+80& =0 36\cdot 0-16x+80& =0 -16x+80& =0 -16x& =-80 x& =\blued{5} \end{aligned}
36yβ16x+80
36β
0β16x+80
β16x+80
β16x
x
β
=0
=0
=0
=β80
=5
β
the solution to the system is x=\blued{5}x=5x, equals, start color blued, 5, end color blued, y=\goldd{0}y=0y, equals, start color goldd, 0, end color goldd.