Solve the system by substitution. Check your solution.
[tex]\begin{array}{l}
y=0.8 x+7.2 \\
20 x+32 y=48
\end{array}[/tex]
a. (8,-6)
b. (5,11)
c. (-4,4)
d. (-10,2)
Answer (1)
Substitute y = 0.8 x + 7.2 into the second equation 20 x + 32 y = 48 .
Solve the resulting equation for x : 20 x + 32 ( 0.8 x + 7.2 ) = 48 ⇒ x = − 4 .
Substitute x = − 4 back into the first equation to find y : y = 0.8 ( − 4 ) + 7.2 = 4 .
The solution is ( − 4 , 4 ) , which corresponds to option c: ( − 4 , 4 ) .
Explanation
Understanding the Problem We are given a system of two equations:
y = 0.8 x + 7.2
20 x + 32 y = 48
Our goal is to solve this system using substitution and then check which of the given options is the correct solution.
Substitution We will substitute the expression for y from equation (1) into equation (2):
20 x + 32 ( 0.8 x + 7.2 ) = 48
Solving for x Now, we solve for x :
20 x + 25.6 x + 230.4 = 48
45.6 x = 48 − 230.4
45.6 x = − 182.4
x = 45.6 − 182.4
x = − 4
Solving for y Next, we substitute the value of x back into equation (1) to find the value of y :
y = 0.8 ( − 4 ) + 7.2
y = − 3.2 + 7.2
y = 4
Checking the Solution So, the solution is ( − 4 , 4 ) . Now, let's check if this solution satisfies both equations:
Equation (1): 4 = 0.8 ( − 4 ) + 7.2 ⇒ 4 = − 3.2 + 7.2 ⇒ 4 = 4 (True)
Equation (2): 20 ( − 4 ) + 32 ( 4 ) = 48 ⇒ − 80 + 128 = 48 ⇒ 48 = 48 (True)
Since the solution ( − 4 , 4 ) satisfies both equations, it is the correct solution.
Checking Given Options Alternatively, we can check each of the given options:
a. ( 8 , − 6 ) : Equation (1): − 6 = 0.8 ( 8 ) + 7.2 ⇒ − 6 = 6.4 + 7.2 ⇒ − 6 = 13.6 (False)
b. ( 5 , 11 ) : Equation (1): 11 = 0.8 ( 5 ) + 7.2 ⇒ 11 = 4 + 7.2 ⇒ 11 = 11.2 (False)
c. ( − 4 , 4 ) : Equation (1): 4 = 0.8 ( − 4 ) + 7.2 ⇒ 4 = − 3.2 + 7.2 ⇒ 4 = 4 (True). Equation (2): 20 ( − 4 ) + 32 ( 4 ) = 48 ⇒ − 80 + 128 = 48 ⇒ 48 = 48 (True)
d. ( − 10 , 2 ) : Equation (1): 2 = 0.8 ( − 10 ) + 7.2 ⇒ 2 = − 8 + 7.2 ⇒ 2 = − 0.8 (False)
Only option c satisfies both equations.
Examples
Systems of equations are used in various real-world applications, such as determining the break-even point for a business. For example, if a company's cost function is C ( x ) = 20 x + 1000 and its revenue function is R ( x ) = 30 x , solving the system of equations y = 20 x + 1000 and y = 30 x will give the number of units the company needs to sell to break even. Another application is in mixture problems, where you need to find the amounts of different solutions to mix to obtain a desired concentration.
