Answer:
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, start text, E, q, u, a, t, i, o, n, space, 1, end text, 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, start text, E, q, u, a, t, i, o, n, space, 2, end text, 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 #e07d10, 2, x, end color #e07d10 and \goldD yystart color #e07d10, y, end color #e07d10 are equal. So let's plug in \goldD{2x}2xstart color #e07d10, 2, x, end color #e07d10 for \goldD yystart color #e07d10, y, end color #e07d10 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 #11accd, 8, end color #11accd, comma, start color #1fab54, 16, end color #1fab54, 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 #11accd, 8, end color #11accd, comma, start color #1fab54, 16, end color #1fab54, 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=3xy, equals, 3, x
x =x=x, equals
y =y=y, equals
[Show solution]
Solving for a variable first, then using substitution
Sometimes using substitution is a little bit trickier. Here's another system of equations:
-3x + y = -9~~~~~~~ \gray{\text{Equation 1}}−3x+y=−9 Equation 1minus, 3, x, plus, y, equals, minus, 9, space, space, space, space, space, space, space, start color gray, start text, E, q, u, a, t, i, o, n, space, 1, end text, end color gray
5x + 4y = 32~~~~~~~ \gray{\text{Equation 2}}5x+4y=32 Equation 25, x, plus, 4, y, equals, 32, space, space, space, space, space, space, space, start color gray, start text, E, q, u, a, t, i, o, n, space, 2, end text, end color gray
Notice that neither of these equations are already solved for xxx or yyy. As a result, the first step is to solve for xxx or yyy first. Here's how it goes:
Step 1: Solve one of the equations for one of the variables.
Let's solve the first equation for yyy:
\begin{aligned} -3x + y &= -9 &\gray{\text{Equation 1}} \\\\ -3x + y + \maroonD{3x} &= -9 +\maroonD{3x} &\gray{\text{Add 3x to each side}} \\\\ y &= {-9 +3x} &\gray{\text{}}\end{aligned}
−3x+y
−3x+y+3x
y
=−9
=−9+3x
=−9+3x
Equation 1
Add 3x to each side
Step 2: Substitute that equation into the other equation, and solve for xxx.
\begin{aligned} 5x + 4\goldD y &= 32 &\gray{\text{Equation 2}} \\\\ 5x +4(\goldD{-9 + 3x}) &= 32 &\gray{\text{Substitute -9 + 3x for y}} \\\\ 5x -36 +12x &= 32 &\gray{\text{}} \\\\ 17x - 36 &= 32 &\gray{\text{}} \\\\ 17x &= 68 &\gray{\text{}} \\\\ \blueD x &\blueD= \blueD4 &\gray{\text{Divide each side by 17}}\end{aligned}
5x+4y
5x+4(−9+3x)
5x−36+12x
17x−36
17x
x
=32
=32
=32
=32
=68
=4
Equation 2
Substitute -9 + 3x for y
Divide each side by 17
Step 3: Substitute x = 4x=4x, equals, 4 into one of the original equations, and solve for yyy.
\begin{aligned} -3\blueD x + y &= -9 &\gray{\text{The first equation}} \\\\ -3(\blueD{4}) +y &= -9 &\gray{\text{Substitute 4 for x}} \\\\ -12 + y &= -9 &\gray{\text{}} \\\\ \greenD y &\greenD= \greenD3 &\gray{\text{Add 12 to each side}} \end{aligned}
hope this helps