# Solve and Substitute (Substitution)

It's easy to solve an equation with one variable: you isolate the variable and find what number it is equal to. But, when you have equations with two variables, it can be impossible to find a single numerical value for either variable. However, if you have two equations, with the same variables, you can often solve for a variable in one equation and then substitute that answer into the other equation to solve. Remember, in math, you can always substitute variables, terms, and numbers for each other if they are equal.

Sometimes, when you are given two equations, you can solve one to give a variable a numerical value, and then just plug that number into the other equation:

*Example*: If $3y=12$ and $3x+2y=20$ what is the value of $x$?

$$\require{cancel}\eqalign{3y&=12\qquad&&\text{If any equation has just one variable, solve that one first}\\\div 3 &\quad \div 3&&\text{Divide both sides by 3}\\y&=4&&\text{Now you know that y=4}\\\\3x+2y&=20&&\text{Take the other equation}\\3x+2(4)&=20&&\text{Replace y with 4}\\3x+8&=20\\-8&\;\;-8&&\text{Subtract 8 from each side}\\3x&=12\\\div3&\quad \div 3&&\text{Divide each side by 3}\\x&=4}$$

Through solving the first equation, you can plug the solution into the second solution and find $x=4$.

Othertimes, both equations contain a variable. In those cases, your goal is to solve one equation "in terms of" one of the variables, and then plug that into the other equation, so that that equation contains only one variable.

*Example*: If $5x+2y=15$ and $3x+2y=18$ what is the value of $x$?

$$\eqalign{5x+2y&=15&&\text{Solve the first equation in terms of one variable (let's do y)}\\-5x\quad&\;\;-5x&&\text{Subtract 5x from each side}\\2y&=15-5x\\\div 2&\quad \div 2&&\text{Divide both sides by 2 to sove for y}\\y&=\dfrac{15-5x}{2}&&\text{Now you know that y=}\dfrac{15-5x}{2}\\\\3x+2y&=18&&\text{Take the other equation}\\3x+2(\dfrac{15-5x}{2})&=18&&\text{Replace y with }\dfrac{15-5x}{2}\\3x+15-5x&=18&&\text{when you multiply }\dfrac{15-5x}{2} \text{ times 2, it cancels out both 2s}\\8x+15&=18&&\text{Combine like terms}\\-15&\;\;-15&&\text{Subtract 15 from both sides}\\8x&=3\\\div8&\;\;\div8&&\text{Divide each side by 8}\\x&=\dfrac{3}{8}}$$

Overall, if you have two equations with two variables, you can always solve one equation (either for a number or for one variable in terms of another) and substitute into the other equation to solve.

#### Practice Problems:

## Solve and Substitute (Substitution)

Solve the following systems of equations with substitution:

$\eqalign{2x&=4y\\3x+5&=y}$

$\eqalign{x&=2y\\-3x+8y&=12}$

$\eqalign{-x+1&=3y\\5x-7&=10y}$

$\eqalign{x-2&=2y\\20x-18&=2y}$

$\eqalign{-2x&=y-8\\-x+15y&=19}$

$\eqalign{.5x&=4y\\2x+2&y=2}$

$\eqalign{\dfrac{1}{2}x&=6y\\2y-5x+5&=15}$

$\eqalign{5x-3&=3y+7\\2x-11&=y+19}$

$\eqalign{21x&=7y\\15x+15y&=29}$

- $\eqalign{-x&=-8y\\-5x+9&=9y}$

#### Answer Key: