What is the solution to the system of equations use the substitution method to solve the system of equations. show your work.

y = x + 5
4x − 4y = −20

Step-by-step explanation:

Let's work to solve this system of equations:

y = 2x \gray{\text{Equation 1}}y=2x        Equation 1

x + y = 24 \gray{\text{Equation 2}}x+y=24        Equation 2

The tricky thing is that there are two variables, xx and yy. If only we could get rid of one of the variables...

Here's an idea! Equation 11 tells us that \goldD{2x}2x and \goldD yy are equal. So let's plug in \goldD{2x}2x for \goldD yy in Equation 22 to get rid of the yy variable in that equation:

\begin{aligned} x + \goldD y &= 24 &\gray{\text{Equation 2}} x + \goldD{2x} &= 24 &\gray{\text{Substitute 2x for y}}\end{aligned}  

x+y

x+2x

=24

=24

Equation 2

Substitute 2x for y

Brilliant! Now we have an equation with just the xx variable that we know how to solve:

x+2x3x 3x3x=24=24=243=8Divide each side by 3

Nice! So we know that xx equals 88. But remember that we are looking for an ordered pair. We need a yy value as well. Let's use the first equation to find yy when xx equals 88:

\begin{aligned} y &= 2\blueD x &\gray{\text{Equation 1}} y &= 2(\blueD8) &\gray{\text{Substitute 8 for x}} \greenD y &\greenD= \greenD{16}\end{aligned}  

y

y

y

=2x

=2(8)

=16

Equation 1

Substitute 8 for x

Sweet! So the solution to the system of equations is (\blueD8, \greenD{16})(8,16). It's always a good idea to check the solution back in the original equations just to be sure.

Let's check the first equation:

\begin{aligned} y &= 2x \greenD{16} &\stackrel?= 2(\blueD{8}) &\gray{\text{Plug in x = 8 and y = 16}} 16 &= 16 &\gray{\text{Yes!}}\end{aligned}  

y

16

16

=2x

=

?

2(8)

=16

Plug in x = 8 and y = 16

Yes!

Let's check the second equation:

\begin{aligned} x +y &= 24 \blueD{8} + \greenD{16} &\stackrel?= 24 &\gray{\text{Plug in x = 8 and y = 16}} 24 &= 24 &\gray{\text{Yes!}}\end{aligned}  

x+y

8+16

24

=24

=

?

24

=24

Plug in x = 8 and y = 16

Yes!

Great! (\blueD8, \greenD{16})(8,16) is indeed a solution. We must not have made any mistakes.

Your turn to solve a system of equations using substitution.

Use substitution to solve the following system of equations.

4x + y = 284x+y=28

y = 3xy=3x

a) 0

step-by-step explanation:

i^413 x i^0 = 0 as any number multiplied by a zero is

happy to

pls mark as brainliest.