Which expression can you substitute in the indicated equation to solve the system below?
x + y = 6
12x + y = 5
1. 6 - y for x in 12x + y = 5
2. 5 + 12x for y in x + y = 6
3. 6 + y for x in 12x + y = 5
4. 5 - x for y in x + y = 6

4x+y4x+3x 7xx=28=28=28=4Substitue 3x for y

Now let's find yyy:

yy y=3x=3(4)=12

The solution:

x = 4x=4x, equals, 4

y = 12y=12

Step-by-step explanation:

Let's work to solve this system of equations:

y = 2x \gray{\text{Equation 1}}y=2x        Equation 1y, equals, 2, x, space, space, space, space, space, space, space, space, start color gray, E, q, u, a, t, i, o, n, space, 1, end color gray

x + y = 24 \gray{\text{Equation 2}}x+y=24        Equation 2x, plus, y, equals, 24, space, space, space, space, space, space, space, space, start color gray, E, q, u, a, t, i, o, n, space, 2, end color gray

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

Here's an idea! Equation 111 tells us that \goldD{2x}2xstart color goldD, 2, x, end color goldD and \goldD yystart color goldD, y, end color goldD are equal. So let's plug in \goldD{2x}2xstart color goldD, 2, x, end color goldD for \goldD yystart color goldD, y, end color goldD in Equation 222 to get rid of the yyy 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 xxx variable that we know how to solve:

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

Nice! So we know that xxx equals 888. But remember that we are looking for an ordered pair. We need a yyy value as well. Let's use the first equation to find yyy when xxx equals 888:

\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)left parenthesis, start color blueD, 8, end color blueD, comma, start color greenD, 16, end color greenD, right parenthesis. 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)left parenthesis, start color blueD, 8, end color blueD, comma, start color greenD, 16, end color greenD, right parenthesis 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=284, x, plus, y, equals, 28

y = 3xy=3x

i think that you should ask google for the answer to that problem because if we don't know it google does

step-by-step explanation: