# SOLVING PROBLEMS WITH SYSTEMS OF EQUATIONS

## About "Solving problems with systems of equations"

Solving problems with systems of equations :

In this section, we are going to see, how real-world problems can be solved using system of equations.

## Solving problems with systems of equations - Examples

Example 1 :

A park charges \$10 for adults and \$5 for kids. How many many adults tickets and kids tickets were sold, if a total of 548 tickets were sold for a total of \$3750 ?

Solution :

Step 1 :

Let "x" be the number of adults tickets and "y" be the number of kids tickets.

No. of adults tickets + No. of kids tickets  =  Total

x + y  =  548 -------- (2)

Step 2 :

Write an equation which represents the total cost.

Cost of "x" no. adults tickets  =  10x

Cost of "y" no. of kids tickets  =  5y

Total cost  =  \$3750

Then, we have

10x + 5y  =  3750

Divide both sides by 5.

2x + y  =  750 -------- (2)

Step 3 :

Solve an equation for one variable.

Select one of the equation, say x + y  =  548.

Solve for the variable y in terms of x.

Subtract x from both sides.

(x + y) - x  =  (548) - x

y  =  548 - x

Step 4 :

Substitute the expression for y in the other equation and solve.

2x + y  =  750

2x + (548 - x)  =  750

Combine like terms.

x + 548  =  750

Subtract 548 from both sides.

x  =  202

Step 5 :

Substitute the value of x we got above (x = 202) into one of the equations and solve for the other variable, y.

x + y  =  548

202 + y  =  548

Subtract 202 from both sides.

y  =  346

So, the solution of the system is (202, 346).

Step 6 :

Interpret the solution in the original context.

Hence, the number of adults tickets sold is 202 and the number of kids tickets sold is 346.

Example 2 :

Hertz Car Rental rents cars for x dollars per day plus y dollars for each mile driven. John rented a car for 4 days, drove it 160 miles, and spent \$120. David rented a car for 1 day, drove it 240 miles, and spent \$80. Write equations to represent John expenses and David expenses. Then solve the system and tell what each number represents.

Solution :

Step 1 :

John rented a car for 4 days, drove it 160 miles, and spent \$120.

So, we have

4x + 160y  =  120

Step 2 :

David rented a car for 1 day, drove it 240 miles, and spent \$80.

So, we have

x + 240y  =  80

Step 3 :

Solve an equation for one variable.

Select one of the equation, say x + 240y  =  80.

Solve for the variable x in terms of y.

Subtract 240y from both sides.

(x + 240y) - 240y  =  (80) - 240y

x  =  80 - 240y

Step 4 :

Substitute the expression for x in the other equation and solve.

4(80 - 240y) + 160y  =  120

320 - 960y + 160y  =  120

Combine like terms.

320 - 800y  =  120

Subtract 320 from both sides.

-800y  =  -200

Divide both sides by -800

(-800y)/(-800)  =  (-200)/(-800)

y  =  0.25

Step 5 :

Substitute the value of y we got above (y = 0.25) into one of the equations and solve for the other variable, y.

x + y  =  548

202 + y  =  548

Subtract 202 from both sides.

x + 240y  =  80

x + 240(0.25)  =  80

x + 60  =  80

Subtract 60 from both sides.

x  =  20

So, the solution of the system is (20, 0.25).

Step 6 :

Interpret the solution in the original context.

Hence, Hertz Car Rental rents cars for \$20 per day plus \$0.25 for each mile driven. After having gone through the stuff given above, we hope that the students would have understood "Solving problems with systems of equations".

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