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# 7.2: Solving Systems by Substitution

Difficulty Level: At Grade Created by: CK-12

While the graphical approach to solving systems is helpful, it may not always provide exact answers. Therefore, we will learn a second method to solving systems. This method uses the Substitution Property of Equality.

Substitution Property of Equality: If y=\begin{align*}y=\end{align*} an algebraic expression, then the algebraic expression can be substituted for any y\begin{align*}y\end{align*} in an equation or an inequality.

Consider the racing example from the previous lesson.

Peter and Nadia like to race each other. Peter can run at a speed of 5 feet per second and Nadia can run at a speed of 6 feet per second. To be a good sport, Nadia likes to give Peter a head start of 20 feet. How long does Nadia take to catch up with Peter? At what distance from the start does Nadia catch up with Peter?

The two racers' information was translated into two equations.

Peter: d=5t+20\begin{align*}d=5t+20\end{align*}

Nadia: d=6t\begin{align*}d=6t\end{align*}

We want to know when the two racers will be the same distance from the start. This means we can set the two equations equal to each other.

5t+20=6t\begin{align*}5t+20=6t\end{align*}

Now solve for t\begin{align*}t\end{align*}.

5t5t+2020=6t5t=1t\begin{align*}5t-5t+20& =6t-5t\\ 20& = 1t\end{align*}

After 20 seconds, Nadia will catch Peter.

Now we need to determine how far from the distance the two runners are. You already know 20=t\begin{align*}20=t\end{align*}, so we will substitute to determine the distance. Using either equation, substitute the known value for t\begin{align*}t\end{align*} and find d\begin{align*}d\end{align*}.

d=5(20)+20120\begin{align*}d=5(20)+20 \rightarrow 120\end{align*}

When Nadia catches Peter, the runners are 120 feet from the starting line.

The Substitution Method is useful when one equation of the system is of the form y=\begin{align*}y=\end{align*} algebraic expression or x=\begin{align*}x=\end{align*} algebraic expression.

Example 1: Solve the system {x+y=2y=3\begin{align*}\begin{cases} x+y=2\\ \qquad y=3 \end{cases}\end{align*}

Solution: The second equation is solved for the variable x\begin{align*}x\end{align*}. Therefore, we can substitute the value “3” for any y\begin{align*}y\end{align*} in the system.

x+y=2x+3=2\begin{align*}x+y=2 \rightarrow x+3=2\end{align*}

Now solve the equation for x:\begin{align*}x:\end{align*} x+33x=23=1\begin{align*}x+3-3& =2-3\\ x & = -1\end{align*}

The x\begin{align*}x-\end{align*}coordinate of the intersection of these two equations is –1. Now we must find the y\begin{align*}y-\end{align*}coordinate using substitution.

x+y1+1+yy=2(1)+y=2=2+1=3\begin{align*}x+y & = 2 \rightarrow (-1)+y=2\\ -1+1+y& = 2+1\\ y & = 3\end{align*}

As seen in the previous lesson, the solution to the system is (–1, 3).

Example: Find the solution to the system {y=3x5y=2x+5\begin{align*}\begin{cases} y=3x-5\\ y=-2x+5 \end{cases}\end{align*} using substitution.

Solution: Each equation is equal to the variable y\begin{align*}y\end{align*}, therefore the two algebraic expressions must equal each other.

3x5=2x+5\begin{align*}3x-5=-2x+5\end{align*}

Solve for x\begin{align*}x\end{align*}.

3x5+53x+2x5xx=2x+5+5=2x+2x+10=10=2\begin{align*}3x-5+5&=-2x+5+5\\ 3x+2x&=-2x+2x+10\\ 5x&=10\\ x&=2\end{align*}

The x\begin{align*}x-\end{align*}coordinate of the intersection of the two lines is 2. Now you must find the y\begin{align*}y-\end{align*}coordinate using either of the two equations.

y=2(2)+5=1\begin{align*}y=-2(2)+5=1\end{align*}

The solution to the system is x=2, y=1\begin{align*}x=2, \ y=1\end{align*} or (2, 1).

## Solving Real-World Systems by Substitution

Example: Anne is trying to choose between two phone plans. Vendaphone’s plan costs $20 per month, with calls costing an additional 25 cents per minute. Sellnet’s plan charges$40 per month, but calls cost only 8 cents per minute. Which should she choose?

Solution: Anne’s choice will depend upon how many minutes of calls she expects to use each month. We start by writing two equations for the cost in dollars in terms of the minutes used. Since the number of minutes is the independent variable, it will be our x\begin{align*}x\end{align*}. Cost is dependent on minutes. The cost per month is the dependent variable and will be assigned y\begin{align*}y\end{align*}.

For VendafoneFor Sellnety=0.25x+20y=0.08x+40\begin{align*}& \text{For Vendafone} && y =0.25x+20\\ & \text{For Sellnet} && y = 0.08x+40\end{align*}

By graphing two equations, we can see that at some point the two plans will charge the same amount, represented by the intersection of the two lines. Before this point, Sellnet’s plan is more expensive. After the intersection, Sellnet’s plan is cheaper.

Use substitution to find the point that the two plans are the same. Each algebraic expression is equal to y\begin{align*}y\end{align*}, so they must equal each other.

0.25x+200.25x0.17xx=0.08x+40=0.08x+20=20=117.65 minutesSubtract 20 from both sides.Subtract 0.08x from both sides.Divide both sides by 0.17.Rounded to two decimal places.\begin{align*}0.25x+20 & =0.08x+40 && \text{Subtract 20 from both sides.}\\ 0.25x & = 0.08x+20 && \text{Subtract} \ 0.08x \ \text{from both sides.}\\ 0.17x & = 20 && \text{Divide both sides by 0.17.}\\ x & = 117.65 \ \text{minutes} && \text{Rounded to two decimal places.}\end{align*}

We can now use our sketch, plus this information, to provide an answer. If Anne will use 117 minutes or fewer every month, she should choose Vendafone. If she plans on using 118 or more minutes, she should choose Sellnet.

## Mixture Problems

Systems of equations arise in chemistry when mixing chemicals in solutions and can even be seen in things like mixing nuts and raisins or examining the change in your pocket!

By rearranging one sentence in an equation into y=\begin{align*}y=\end{align*} algebraic expression or x=\begin{align*}x=\end{align*} algebraic expression, you can use the Substitution Method to solve the system.

Example: Nadia empties her purse and finds that it contains only nickels and dimes. If she has a total of 7 coins and they have a combined value of 55 cents, how many of each coin does she have?

Solution: Begin by choosing appropriate variables for the unknown quantities. Let n=\begin{align*}n=\end{align*} the number of nickels and d=\begin{align*}d=\end{align*} the number of dimes.

There are seven coins in Nadia’s purse: n+d=7\begin{align*}n+d=7\end{align*}.

The total is 55 cents: 0.05n+0.10d=0.55\begin{align*}0.05n+0.10d=0.55\end{align*}.

The system is: {  n+d=70.05n+0.10d=0.55\begin{align*}\begin{cases} \qquad \quad \ \ n+d=7\\ 0.05n+0.10d=0.55 \end{cases}\end{align*}.

We can quickly rearrange the first equation to isolate d\begin{align*}d\end{align*}, the number of dimes: d=7n\begin{align*}d=7-n\end{align*}.

Using the Substitution Property, every d\begin{align*}d\end{align*} can be replaced with the expression 7n\begin{align*}7-n\end{align*}.

0.05n+0.10(7n)Now solve for n:0.05n+0.700.10n0.05n+0.700.05nn=0.55=0.55Distributive Property=0.55Add like terms.=0.15 Subtract 0.70.=3 Divide by 0.05.\begin{align*}0.05n+0.10(7-n)&=0.55\\ \text{Now solve for} \ n: \qquad 0.05n+0.70-0.10n&=0.55 \qquad \text{Distributive Property}\\ -0.05n+0.70&=0.55 \qquad \text{Add like terms.}\\ -0.05n&=-0.15 \quad \ \text{Subtract} \ 0.70.\\ n& =3 \qquad \quad \ \text{Divide by} \ -0.05.\end{align*}

Nadia has 3 nickels. There are seven coins in the purse; three are nickels so four must be dimes.

Check to make sure this combination is 55 cents: 0.05(3)+0.10(4)=0.15+0.40=0.55\begin{align*}0.05(3)+ 0.10(4)= 0.15+0.40=0.55\end{align*}.

## Chemical Mixtures

Example: A chemist has two containers, Mixture A\begin{align*}A\end{align*} and Mixture B\begin{align*}B\end{align*}. Mixture A\begin{align*}A\end{align*} has a 60% copper sulfate concentration. Mixture B\begin{align*}B\end{align*} has a 5% copper sulfate concentration. The chemist needs to have a mixture equaling 500 mL with a 15% concentration. How much of each mixture does the chemist need?

Solution: Although not explicitly stated, there are two equations involved in this situation.

• Begin by stating the variables. Let A=mixture A and B=mixture B\begin{align*}A = mixture \ A \ and \ B = mixture \ B\end{align*}.
• The total mixture needs to have 500 mL of liquid.

Equation 1 (how much total liquid): A+B=500\begin{align*}A+B=500\end{align*}.

• The total amount of copper sulfate needs to be 15% of the total amount of solution (500 mL). 0.15500=75 ounces\begin{align*}0.15 \cdot 500=75 \ ounces\end{align*}

Equation 2 (how much copper sulfate the chemist needs): 0.60A+0.05B=75\begin{align*}0.60A+0.05B=75\end{align*}

{A+B=5000.60A+0.05B=75\begin{align*}\begin{cases} A+B=500\\ 0.60A+0.05B=75 \end{cases}\end{align*}

By rewriting equation 1, the Substitution Property can be used: A=500B\begin{align*}A=500-B\end{align*}.

Substitute the expression 500B\begin{align*}500-B\end{align*} for the variable A\begin{align*}A\end{align*} in the second equation.

0.60(500B)+0.05B=75\begin{align*}0.60(500-B)+0.05B=75\end{align*}

Solve for B\begin{align*}B\end{align*}.

3000.60B+0.05B3000.55B0.55BB=75=75=225409 mLDistributive PropertyAdd like terms.Subtract 300.\begin{align*}300-0.60B+0.05B& =75 && \text{Distributive Property}\\ 300-0.55B&=75 && \text{Add like terms.}\\ -0.55B&=-225 && \text{Subtract} \ 300.\\ B & \approx 409 \ mL\end{align*}

The chemist needs approximately 409 mL of mixture B\begin{align*}B\end{align*}. To find the amount of mixture A\begin{align*}A\end{align*}, use the first equation: A+409=500\begin{align*}A+409=500\end{align*}

A=91 mL\begin{align*}A=91 \ mL\end{align*}

The chemist needs 91 milliliters of mixture A\begin{align*}A\end{align*} and 409 milliliters of mixture B\begin{align*}B\end{align*} to get a 500 mL solution with a 15% copper sulfate concentration.

## Practice Set

Sample explanations for some of the practice exercises below are available by viewing the following video. Note that there is not always a match between the number of the practice exercise in the video and the number of the practice exercise listed in the following exercise set.  However, the practice exercise is the same in both.

1. Explain the process of solving a system using the Substitution Property.
2. Which systems are easier to solve using substitution?

Solve the following systems. Remember to find the value for both variables!

1. {y=36x2y=0\begin{align*}\begin{cases} y=-3\\ 6x-2y=0 \end{cases}\end{align*}
2. {33y=6y=3x+4\begin{align*}\begin{cases} -3-3y=6\\ y=-3x+4 \end{cases}\end{align*}
3. {y=3x+16y=x+8\begin{align*}\begin{cases} y=3x+16\\ y=x+8 \end{cases}\end{align*}
4. {y=6x3y=3\begin{align*}\begin{cases} y=-6x-3\\ y=3 \end{cases}\end{align*}
5. {y=2x+5y=18x\begin{align*}\begin{cases} y=-2x+5\\ y=-1-8x \end{cases}\end{align*}
6. {y=6+xy=2x15\begin{align*}\begin{cases} y=6+x\\ y=-2x-15 \end{cases}\end{align*}
7. {y=2y=5x17\begin{align*}\begin{cases} y=-2\\ y=5x-17 \end{cases}\end{align*}
8. {x+y=53x+y=15\begin{align*}\begin{cases} x+y=5\\ 3x+y=15 \end{cases}\end{align*}
9. {12y3x=1x4y=1\begin{align*}\begin{cases} 12y-3x=-1\\ x-4y=1 \end{cases}\end{align*}
10. x+2y=93x+5y=20\begin{align*}x+2y=9\!\\ 3x+5y=20 \end{align*}
11. x3y=102x+y=13\begin{align*}x-3y=10\!\\ 2x+y=13 \end{align*}
12. Solve the system \begin{align*}\begin{cases} y=\frac{1}{4} x-14\\ y=\frac{19}{8} x+7 \end{cases}\end{align*} by graphing and substitution. Which method do you prefer? Why?
13. Of the two non-right angles in a right angled triangle, one measures twice that of the other. What are the angles?
14. The sum of two numbers is 70. They differ by 11. What are the numbers?
15. A rectangular field is enclosed by a fence on three sides and a wall on the fourth side. The total length of the fence is 320 yards. If the field has a total perimeter of 400 yards, what are the dimensions of the field?
16. A ray cuts a line forming two angles. The difference between the two angles is \begin{align*}18^\circ\end{align*}. What does each angle measure?
17. I have $15.00 and wish to buy five pounds of mixed nuts for a party. Peanuts cost$2.20 per pound. Cashews cost $4.70 per pound. How many pounds of each should I buy? 18. A chemistry experiment calls for one liter of sulfuric acid at a 15% concentration, but the supply room only stocks sulfuric acid in concentrations of 10% and in 35%. How many liters of each should be mixed to give the acid needed for the experiment? 19. Bachelle wants to know the density of her bracelet, which is a mix of gold and silver. Density is total mass divided by total volume. The density of gold is 19.3 g/cc and the density of silver is 10.5 g/cc. The jeweler told her that the volume of silver used was 10 cc and the volume of gold used was 20 cc. Find the combined density of her bracelet. 20. Jeffrey wants to make jam. He needs a combination of raspberries and blackberries totaling six pounds. He can afford$11.60. How many pounds of each berry should he buy?

Mixed Review

1. The area of a square is \begin{align*}96 \ inches^2\end{align*}. Find the length of a square exactly.
2. The volume of a sphere is \begin{align*}V= \frac{4}{3} \pi r^3\end{align*}, where \begin{align*}r=radius\end{align*}. Find the volume of a sphere with a diameter of 11 centimeters.
3. Find:
2. the multiplicative inverse of 7.6.
4. Solve for \begin{align*}x: \frac{1.5}{x}=6\end{align*}.
5. The temperature in Fahrenheit can be approximated by crickets using the rule “Count the number of cricket chirps in 15 seconds and add 40.”
1. What is the domain of this function?
2. What is the range?
3. Would you expect to hear any crickets at \begin{align*}32^\circ F\end{align*}? Explain your answer.
4. How many chirps would you hear if the temperature were \begin{align*}57^\circ C\end{align*}?
6. Is 4.5 a solution to \begin{align*}45-6x \le 18\end{align*}?

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