# Systems Using Substitution

## Solve for one variable, substitute the value in the other equation

%
Progress

MEMORY METER
This indicates how strong in your memory this concept is
Progress
%
Solving Systems with One Solution Using Substitution
Rex and Carl are making a mixture in science class. They need to have 12 ounces of a 60% saline solution. To make this solution they have a 20% saline solution and an 80% saline solution. How many ounces of each do they need to make the correct mixture?

### Systems with One Solution Using Substitution

In the substitution method we will be looking at the two equations and deciding which variable is easiest to solve for so that we can write one of the equations as \begin{align*}x=\end{align*} or \begin{align*}y=\end{align*}. Next we will replace either the \begin{align*}x\end{align*} or the \begin{align*}y\end{align*} accordingly in the other equation. The result will be a solve-able equation with only one variable.

Let's solve the systems below using substitution.

1.

\begin{align*}2x+y &= 12\\ \text{-}3x-5y &= \text{-}11\end{align*}

The first step is to look for a variable that is easy to isolate. In other words, does one of the variables have a coefficient of 1? Yes, that variable is the \begin{align*}y\end{align*} in the first equation. So, start by solving for (isolating) \begin{align*}y: \ \ y=\text{-}2x+12.\end{align*}

This expression can then be used to replace the \begin{align*}y\end{align*} in the other equation to solve for \begin{align*}x:\end{align*}

\begin{align*}\text{-}3x-5(\text{-}2x+12) &= \text{-}11\\ \text{-}3x+10x-60 &= \text{-}11\\ 7x-60 &= \text{-}11\\ 7x &= 49\\ x &= 7\end{align*}

Now that we have found \begin{align*}x\end{align*}, we can use this value in our expression to find \begin{align*}y:\end{align*}

\begin{align*} y &= \text{-}2(7)+12\\ y &= \text{-}14+12\\ y &= \text{-}2\end{align*}

Recall that the solution to a linear system is a point in the coordinate plane where the two lines intersect. Therefore, our answer should be written as a point: (7, -2). You can check your answer by substituting this point into both equations to make sure that it satisfies them:

\begin{align*}2(7)+ \text{-}2 &= 14-2=12\\ \text{-}3(7)-5(\text{-}2) &= \text{-}21+10 = \text{-}11 \end{align*}

1.

\begin{align*}2x+3y &= 13\\ x+5y &= \text{-}4\end{align*}

In the last problem, \begin{align*}y\end{align*} was the easiest variable to isolate. Is that the case here? No, this time, \begin{align*}x\end{align*} is the variable with a coefficient of 1. It is easy to fall into the habit of always isolating \begin{align*}y\end{align*} since you have done it so much to write equation in slope-intercept form. Try to avoid this and look at each system to see which variable is easiest to isolate. Doing so will help reduce your work.

Solving the second equation for \begin{align*}x\end{align*} gives: \begin{align*}x = \text{-}5y-4.\end{align*}

This expression can be used to replace the \begin{align*}x\end{align*} in the other equation and solve for \begin{align*}y:\end{align*}

\begin{align*}2(\text{-}5y-4)+3y &= 13\\ \text{-}10y-8+3y &= 13\\ \text{-}7y-8 &= 13\\ \text{-}7y &= 21\\ y &= \text{-}3\end{align*}

Now that we have found \begin{align*}y\end{align*}, we can use this value in our expression to find \begin{align*}x\end{align*}:

\begin{align*}x &= \text{-}5(\text{-}3)-4\\ x &= 15-4\\ x &= 11\end{align*}

So, the solution to this system is (11, -3). Don’t forget to check your answer:

\begin{align*}2(11)+3(\text{-}3) &= 22-9=13\\ 11+5(\text{-}3) &= 11-15= \text{-}4 \end{align*}

1.

\begin{align*}4x+3y &=4\\ 6x-2y &= 19\end{align*}

In this case, none of the variables have a coefficient of 1. So let’s solve for the \begin{align*}x\end{align*} in equation 1:

\begin{align*}4x &= \text{-}3y +4\\ x &= \text{-} \frac{3}{4}y+1\end{align*}

Now this expression can be used to replace the \begin{align*}x\end{align*} in the other equation and solve for \begin{align*}y:\end{align*}

\begin{align*}6 \left(\text{-} \frac{3}{4}y+1 \right)-2y &= 19\\ \text{-} \frac{18}{4}y+6-2y &= 19\\ \text{-} \frac{9}{2}y- \frac{4}{2}y &= 13\\ \text{-} \frac{13}{2}y &= 13\\ \left(\text{-}\frac{2}{13} \right)\left( \text{-} \frac{13}{2} \right)y &= 13\left( \text{-} \frac{2}{13} \right)\\ y &= \text{-}2\end{align*}

Now that we have found \begin{align*}y\end{align*}, we can use this value in our expression find \begin{align*}x:\end{align*}

\begin{align*}x &= \left(- \frac{3}{4} \right)(-2)+1\\ x &= \frac{6}{4}+1 \\ x &= \frac{3}{2}+\frac{2}{2}\\ x &= \frac{5}{2}\end{align*}

So, the solution is \begin{align*}\left( \frac{5}{2}, \text{-}2 \right)\end{align*}. Check your answer:

\begin{align*}4\left(\frac{5}{2}\right)+3(\text{-}2) &= 10-6=4\\ 6\left(\frac{5}{2}\right)-2(\text{-}2) &= 15+4 =19 \end{align*}

### Examples

#### Example 1

Earlier, you were asked to find the number of ounces of each of two concentrations of solution Rex and Carl need to make a particular mixture.

Let’s try to make the word problem easier by organizing the information into a “picture” equation as shown below:

In this picture, we can see that they will be mixing \begin{align*}x\end{align*} ounces of the 20% solution with \begin{align*}y\end{align*} ounces of the 80% solution to get 12 ounces of the 60% solution. The two equations are thus:

\begin{align*}0.2x+0.8y &= 0.6(12)\\ x+y &= 12\end{align*}

Now we can solve the system using substitution. Solve for \begin{align*}y\end{align*} in the second equation to get \begin{align*}y = 12-x.\end{align*}

Now, substitute and solve in the first equation:

\begin{align*}0.2x+0.8(12-x) &= 0.6(12)\\ 0.2x+9.6-0.8x &= 7.2\\ \text{-}0.6x &= \text{-}2.4\\ x &= 4\end{align*}

Now find \begin{align*}y\end{align*}:

\begin{align*}y &= 12-x\\ y &= 12-4\\ y &= 8\end{align*}

Therefore, Rex and Carl need 4 ounces of the 20% saline solution and 8 ounces of the 80% saline solution to make the correct mixture.

Solve the following systems using the substitution method.

#### Example 2

\begin{align*}3x+4y &= \text{-}13\\ x &=\text{-}2y-9\end{align*}

In this problem, the second equation is already solved for \begin{align*}x,\end{align*} so we can use that in the first equation to find \begin{align*}y\end{align*}:

\begin{align*}3(\text{-}2y-9)+4y &= \text{-}13\\ \text{-}6y-27+4y &= \text{-}13\\ \text{-}2y-27 &= \text{-}13\\ \text{-}2y &= 14\\ y &= \text{-}7\end{align*}

Now we can find \begin{align*}x\end{align*}:

\begin{align*}x &= \text{-}2(\text{-}7)-9\\ x &= 14-9\\ x &= 5\end{align*}

Therefore the solution is (5, -7).

#### Example 3

\begin{align*}\text{-}2x-5y &= \text{-}39\\ x+3y &=24\end{align*}

This time the \begin{align*}x\end{align*} in the second equation is the easiest variable to isolate: \begin{align*}x=-3y+24.\end{align*} Let’s use this in the first expression to find \begin{align*}y\end{align*}:

\begin{align*}\text{-}2(\text{-}3y+24)-5y &= \text{-}39\\ 6y-48-5y &= \text{-}39\\ y-48 &= \text{-}39\\ y &=9\end{align*}

Now we can find \begin{align*}x\end{align*}:

\begin{align*}x &= \text{-}3(9)+24\\ x &= \text{-}27+24\\ x &= \text{-}3\end{align*}

Therefore the solution is (-3, 9).

#### Example 4

\begin{align*}y &= \frac{1}{2}x-21\\ y &= \text{-}2x+9\end{align*}

In this case, both equations are equal to \begin{align*}y\end{align*}. Since \begin{align*}y=y\end{align*}, by the Reflexive Property of Equality, we can set the right hand sides of the equations be equal too. This is still a substitution problem; it just looks a little different.

\begin{align*}\frac{1}{2}x-21 &= \text{-}2x+9\\ 2 \left( \frac{1}{2}x-21 \right) &= 2(\text{-}2x+9)\\ x-42 &= \text{-}4x+18\\ 5x &= 60\\ x &= 12\end{align*}

Now we can find \begin{align*}y\end{align*}:

\begin{align*}y &= \frac{1}{2}(12)-21 \qquad \quad \quad y = \text{-}2(12)+9\\ y &= 6-21 \qquad \quad \text{or} \qquad y = \text{-}24+9\\ y &= \text{-}15 \qquad \qquad \qquad \quad y = \text{-}15\end{align*}

Therefore, the solution is (12, -15).

### Review

1. . \begin{align*}\begin{cases} x+3y= \text{-}1\\ 2x+9y= 7 \end{cases}\end{align*}
1. .a\begin{align*}\ \\ \begin{cases} 5x+2y= 0\\ y= x-7 \end{cases}\\ \ \\\end{align*}
1. .\begin{align*}\begin{cases} 5x+2y= 0\\ y= x-7 \end{cases}\end{align*}
1. .\begin{align*}\ \\ \begin{cases} 2x-5y= 21\\ x= \text{-}6y +2 \end{cases}\\ \ \\\end{align*}
1. .\begin{align*}\begin{cases} y= x+3\\ y= 2x-1 \end{cases}\end{align*}
1. .\begin{align*}\ \\ \begin{cases}x+6y= 1\\ \text{-}2x-11y= \text{-}4 \end{cases}\\ \ \\\end{align*}
1. . \begin{align*}\begin{cases}2x+y= 18\\ \text{-}3x+11y= \text{-}27 \end{cases}\end{align*}
1. .\begin{align*}\ \\ \begin{cases}2x+3y= 5\\ 5x+7y= 8 \end{cases}\\ \ \\\end{align*}
1. .\begin{align*}\begin{cases}\text{-}7x+2y= 9\\ 5x-3y= 3 \end{cases}\end{align*}
1. .\begin{align*}\ \\ \begin{cases}2x-6y= \text{-}16\\ \text{-}6x+10y= 8 \end{cases}\\ \ \\\end{align*}
1. .\begin{align*}\begin{cases}2x-3y= \text{-}3\\ 8x+6y= 12 \end{cases}\end{align*}
1. .\begin{align*}\ \\ \begin{cases}5x+y= \text{-}3\\ y= 15x+9\end{cases}\\ \ \\\end{align*}

Set up and solve a system of linear equations to answer each of the following word problems.

1. Alicia and Sarah are at the supermarket. Alicia wants to get peanuts from the bulk food bins and Sarah wants to get almonds. The almonds cost $6.50 per pound and the peanuts cost$3.50 per pound. Together they buy 1.5 pounds of nuts. If the total cost is $6.75, how much did each girl get? Set up a system to solve using substitution. 2. Marcus goes to the department store to buy some new clothes. He sees a sale on t-shirts ($5.25) and shorts ($7.50). Marcus buys seven items and his total, before sales tax, is$43.50. How many of each item did he buy?
3. Jillian is selling tickets for the school play. Student tickets are $3 and adult tickets are$5. If 830 people buy tickets and the total revenue is \$3104, how many students attended the play?

### Notes/Highlights Having trouble? Report an issue.

Color Highlighted Text Notes

### Vocabulary Language: English

TermDefinition
Consistent A system of equations is consistent if it has at least one solution.
distributive property The distributive property states that the product of an expression and a sum is equal to the sum of the products of the expression and each term in the sum. For example, $a(b + c) = ab + ac$.
linear equation A linear equation is an equation between two variables that produces a straight line when graphed.
substitute In algebra, to substitute means to replace a variable or term with a specific value.