The tricky thing is that there are two variables, and . If only we could get rid of one of the variables...
Here's an idea! Equation 1111 tells us that 2x\goldD{2x}2xstart color #e07d10, 2, x, end color #e07d10 and y\goldD yystart color #e07d10, y, end color #e07d10 are equal. So let's plug in 2x\goldD{2x}2xstart color #e07d10, 2, x, end color #e07d10 for y\goldD yystart color #e07d10, y, end color #e07d10 in Equation 2222 to get rid of the variable in that equation:
x+y=24Equation 2x+2x=24Substitute 2x for y\begin{aligned} x + \goldD y &= 24 &\gray{\text{Equation 2}} x + \goldD{2x} &= 24 &\gray{\text{Substitute 2x for y}}\end{aligned}x+yx+2x=24=24Equation 2Substitute 2x for y
Brilliant! Now we have an equation with just the variable that we know how to solve:
x+2x3x 3x3x=24=24=243=8Divide each side by 3
Nice! So we know that equals 8888. But remember that we are looking for an ordered pair. We need a value as well. Let's use the first equation to find when equals 8888:
y=2xEquation 1y=2(8)Substitute 8 for xy=16\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}yyy=2x=2(8)=16Equation 1Substitute 8 for x
Sweet! So the solution to the system of equations is (8,16)(\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:
y=2x16=?2(8)Plug in x = 8 and y = 1616=16Yes!\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}y1616=2x=?2(8)=16Plug in x = 8 and y = 16Yes!
Let's check the second equation:
x+y=248+16=?24Plug in x = 8 and y = 1624=24Yes!\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+y8+1624=24=?24=24Plug in x = 8 and y = 16Yes!
Great! (8,16)(\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 = 284x+y=284, x, plus, y, equals, 28
y=3xy = 3xy=3x
