for each value of y, determine whether it is a solution to 33 = 40 - y.
is it a solution?

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

5y−7x

​  

=−5

=12

​  

we notice that the first equation has a 7x7x7, x term and the second equation has a -7x−7xminus, 7, x term. these terms will cancel if we add the equations together—that is, we'll eliminate the xxx terms:

2y+7x+ 5y−7x7y+0=−5=12=7

solving for yyy, we get:

\begin{aligned} 7y+0 & =7 7y & =7 y & =\goldd{1} \end{aligned}

7y+0

7y

y

​  

=7

=7

=1

​  

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

2⋅1+7x

2+7x

7x

x

​  

=−5

=−5

=−5

=−7

=−1

​  

the solution to the system is x=\blued{-1}x=−1x, 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}

5y−7x

5⋅1−7(−1)

5+7

​  

=12

=

?

12

=12

​  

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}

−9y+4x−20

−7y+16x−80

​  

=0

=0

​  

we can multiply the first equation by -4−4minus, 4 to get an equivalent equation that has a \purpled{-16x}−16xstart 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}

36y−16x+80

−7y+16x−80

​  

=0

=0

​  

adding the equations to eliminate the xxx terms, we get:

36y−16x+80+ −7y+16x−8029y+0−0=0=0=0

solving for yyy, we get:

\begin{aligned} 29y+0 -0& =0 29y& =0 y& =\goldd 0 \end{aligned}

29y+0−0

29y

y

​  

=0

=0

=0

​  

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}

36y−16x+80

36⋅0−16x+80

−16x+80

−16x

x

​  

=0

=0

=0

=−80

=5

​  

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.

well what did you do it matters what you did : )

step-by-step explanation: