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Wize High School Algebra I Textbook (Common Core) > Systems of Linear Equations

Solving by Substitution

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Solving by Substitution


One method for solving a system of linear equations is substitution:
  1. Rewrite one of the equations into x=...x=... or y=...y=...
  2. Then "substitute" this equation into the other equation (replace xx or yy with this new ...... value)
  3. Solve for the variable that's left over in this new quation
  4. Use this answer to go back and solve for the other variable

It sounds more complicated than it is! Let's take a look at an example.
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Example
Let's go through the steps for solving by substitution using this system of linear equations as an example:

2x−3y=1     1◯2x-3y=1~~~~~\text{\textcircled 1}
4x+y=−5     2◯4x+y=-5~~~~~\text{\textcircled 2}

Step 1: Rewrite one equation into x=...\bco {x=...} or y=...\bco {y=...}
Since the equation 2◯\text{\textcircled 2} has a stand-alone yy , it is probably the easiest to manipulate.

4x+y=−5y=−4x−5\begin{array}{rcl} 4x+y&=&-5\\ \colorbox{yellow}{$y$}&\bm=&\colorbox{yellow}{$-4x-5$} \end{array}

Step 2: Substitute this simple equation into the remaining equation
We substitute (or replace) the yy in equation 1◯\text{\textcircled 1} with this new value of yy:
2x−3y=12x−3(−4x−5)=1\begin{array}{rcl} 2x-3\colorbox{yellow}{$y$}&=&1\\\\ 2x-3(\colorbox{yellow}{$-4x-5$})&=&1\\\\ \end{array}

Step 3: Solve for the variable in this new equation
2x+12x+15=114x+15=114x=1−1514x14=−1414x=−1\begin{array}{rcl} 2x+12x+15&=&1\\\\ 14x+15&=&1\\\\ 14x&=&1-15\\\\ \dfrac{14x}{14}&=&\dfrac{-14}{14}\\\\ x&=&-1 \end{array}

Step 4: Use this value to solve for the other variable
Substitute x=−1\colorbox{pink}{$x$}=\colorbox{pink}{$-1$} back into equation y=−4x−5y=-4x-5:
y=−4(−1)−5y=4−5y=−1\begin{array}{rcl} y&=&-4(\colorbox{pink}{$-1$})-5\\ y&=&4-5\\ y&=&-1 \end{array}

Therefore, the solution to this system of linear equations (meaning the point of intersection) is (−1,−1)\boxed{(-1,-1)}.

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(Optional)
We can confirm the solution by putting the solution into the original formulas.

Equation 1: 2x−3y=12x-3y=1
2x−3y=12(−1)−3(−1)=1−2+3=11=1   Confirmed!\begin{array}{rcl} 2x-3y&=&1\\ 2(-1)-3(-1)&=&1\\ -2+3&=&1\\ 1&=&1~~~\text{Confirmed!} \end{array}

Equation 2: 4x+y=−54x+y=-5
4x+y=−54(−1)+(−1)=−5−4−1=−5−5=−5   Confirmed!\begin{array}{rcl} 4x+y=-5\\ 4(-1)+(-1)&=&-5\\ -4-1&=&-5\\ -5&=&-5~~~\text{Confirmed!} \end{array}
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Example: Solving by Substitution

Solve the following system of linear equations:
5x+2y=131◯3x+y=72◯\begin{array}{cc} 5x+2y=13&\text{\textcircled 1}\\ 3x+y=7&\text{\textcircled 2} \end{array}

Step 1:

Let's rearrange equation 2◯\text{\textcircled 2} for the stand-alone yy:
y=7−3xy=7-3x


Step 2:

Substitute y=7−3x\colorbox{yellow}{$y$}=\colorbox{yellow}{$7-3x$} into equation 1◯\text{\textcircled 1}:
5x+2y=135x+2(7−3x)=135x+14−6x=13−x+14=13−x=13−14−x−1=−1−1x=1\begin{array}{rcl} 5x+2\colorbox{yellow}{$y$}&=&13\\\\ 5x+2(\colorbox{yellow}{$7-3x$})&=&13\\\\ 5x+14-6x&=&13\\\\ -x+14&=&13\\\\ -x&=&13-14\\\\ \dfrac{-x}{-1}&=&\dfrac{-1}{-1}\\\\ x&=&1 \end{array}

Step 3:

Use x=1\colorbox{pink}{$x$}=\colorbox{pink}{$1$} to solve for yy in y=7−3xy=7-3x:
y=7−3(x)y=7−3(1)y=7−3y=4\begin{array}{rcl} y&=&7-3(\colorbox{pink}{$x$})\\ y&=&7-3(\colorbox{pink}{$1$})\\ y&=&7-3\\ y&=&4 \end{array}

Therefore, the solution (point of intersection) for the system of linear equations is (1,4)\boxed{(1,4)}.

Let's confirm our solution.

Equation 1: 5x+2y=135x+2y=13
5x+2y=135(1)+2(4)=135+8=1313=13    Confirmed!\begin{array}{rcl} 5x+2y&=&13\\ 5(1)+2(4)&=&13\\ 5+8&=&13\\ 13&=&13~~~\text{ Confirmed!} \end{array}


Equation 2: 3x+y=73x+y=7
3x+y=73(1)+(4)=73+4=77=7    Confirmed!\begin{array}{rcl} 3x+y&=&7\\ 3(1)+(4)&=&7\\ 3+4&=&7\\ 7&=&7~~~\text{ Confirmed!} \end{array}

Practice: Solving by Substitution

Find the point of intersection between the line y=−2x+3y=-2x+3 and y=3x−7y=3x-7 using substitution.

Practice Solving by Substitution

Solve the system of equations.

y=9−x1◯6x−5y−10=02◯\begin{array}{cc} y=9-x&\text{\textcircled 1}\\ 6x-5y-10=0&\text{\textcircled 2} \end{array}

Practice Solving by Substitution

Puneet is a social media influencer who earns money by creating social media posts to help promote products that she believes in. She charges the following to create and post an ad on two different social media platforms:
  • Platform A: $0.10 per "like", plus a $150 flat fee
  • Platform B: $0.15 per "like", plus a $90 flat fee
*A "like" is when a user views the post and then clicks on a "like" button; a flat fee is a fee that she charges no matter how many people views or likes the add.

If Puneet were to post an ad on both platforms A and B, how many "likes" must the ad get for Puneet to make the same amount of money on both platforms? How much money will Puneet make on each platform in this scenario?