# 7.2: Systems Using Substitution

Difficulty Level: Basic Created by: CK-12
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Practice Systems Using Substitution

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Suppose that at a bakery, bagels are sold for one price, and muffins are sold for another price. 4 bagels and 2 muffins cost $11, while 3 bagels and 3 muffins cost$12. How much are bagels and muffins? Could you set up a system of equations to solve for the prices? Do you think you could solve this system by substitution? After completing this Concept, you'll be able to use substitution to solve systems of linear equations so that you can handle scenarios like this.

### Guidance

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 Concept.

#### Example A

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 B

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 C

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.

### Guided Practice

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 Concept, the solution to the system is (–1, 3).

### Practice

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. CK-12 Basic Algebra: Solving Linear Systems by Substitution (9:21)

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+2y3x+5y=9=20\begin{align*}x+2y&=9\\ 3x+5y&=20 \end{align*}
11. x3y2x+y=10=13\begin{align*}x-3y&=10\\ 2x+y&=13 \end{align*}
12. Solve the system {y=14x14y=198x+7\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 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 18\begin{align*}18^\circ\end{align*}. What does each angle measure?

Mixed Review

1. Solve for h:25|2h+5|\begin{align*}h: 25 \ge |2h+5|\end{align*}.
2. Subtract: \begin{align*}\frac{4}{3}-\frac{1}{2}\end{align*}.
3. You write the letters to ILLINOIS on separate pieces of paper and place them into a hat.
1. Find \begin{align*}P\end{align*}(drawing an \begin{align*}I\end{align*}).
2. Find the odds of drawing an \begin{align*}L\end{align*}.
4. Graph \begin{align*}x<2\end{align*} on a number line and on a Cartesian plane.
5. Give an example of an ordered pair in quadrant II.
6. The data below show the average life expectancy in the United States for various years.
1. Use the method of interpolation to find the average life expectancy in 1943.
2. Use the method of extrapolation to find the average life expectancy in 2000.
3. Find an equation for the line of best fit. How do the predictions of this model compare to your answers in questions a) and b)?
U. S. Life Expectancy at Birth
Birth Year Female Male Combined
1940 65.2 60.8 62.9
1950 71.1 65.6 68.2
1960 73.1 66.6 69.7
1970 74.7 67.1 70.8
1975 76.6 68.6 72.6
1980 77.5 70.0 73.7
1985 78.2 71.2 74.7
1990 78.8 71.8 75.4
1995 78.9 72.5 75.8
1998 79.4 73.9 76.7

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### Vocabulary Language: English Spanish

TermDefinition
Substitution Property of Equality If $y=$ an algebraic expression, then the algebraic expression can be substituted for any $y$ in an equation or an inequality.
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.

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