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.

the answer is thin because within context, the stanza concludes the and frail tone the people have toward each other, and the fact that it would be thin, makes the most sense compared to "pliable" or "not hard or firm to the touch". hope this , good luck.

step-by-step explanation:

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