## ELIMINATION METHOD WORKSHEET

Problem 1 :

Solve by elimination method.

3x + 4y  =  -25

2x - 3y  =  6

Problem 2 :

Solve by elimination method

2x + 3y  =  5

3x + 4y  =  7

Problem 3 :

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 ? Problem 1 :

Solve by elimination method.

3x + 4y  =  -25

2x - 3y  =  6

Solution :

3x + 4y  =  -25 ---- (1)

2x - 3y  =  6  ---- (2) Both x terms and y terms have different coefficients in the above system of equations.

Let's try to make the coefficients of y terms equal.

To make the coefficients of y terms equal, we have to find the least common multiple 4 and 3.

The least common multiple of 4 and 3 is 12.

Multiply the first equation by 3 in order to make the coefficient of y as 12 and multiply the second equation by 4 in order to make the coefficient of y as -12.

(1) ⋅ 3 ---->  9x + 12y  =  -75

(2)  4 ---->  8x - 12y  =  24

Now, we can add the two equations and eliminate y as shown below. Divide each side by 17.

x  =  -3

Substitute -3 for x in (1).

(1)---->  3(-3) + 4y  =  -25

-9 + 4y  =  -25

4y  =  -16

Divide each side by 4.

y  =  -4

So, the values of x and y are -3 and -4 respectively.

Problem 2 :

Solve by elimination method

2x + 3y  =  5

3x + 4y  =  7

Solution :

2x + 3y  =  5  ----(1)

3x + 4y  =  7  ----(2) Both x terms and y terms have different coefficients in the above system of equations.

Let's try to make the coefficients of x terms equal.

To make the coefficients of y terms equal, we have to find the least common multiple 2 and 3.

The least common multiple of 2 and 3 is 6.

Multiply the first equation by 3 in order to make the coefficient of x as 6 and multiply the second equation by -2 in order to make the coefficient of x as -6.

(1) ⋅ 3 ---->  6x + 9y  =  15

(2)  -2 ----> -6x - 8y  =  -14

Now, we can add the two equations and eliminate x as shown below. Substitute 1 for y in (1).

(1)---->  2x + 3(1)  =  5

2x + 3  =  5

Subtract 3 from each side.

2x  =  2

Divide each side by 2.

x  =  1

So, the values of x and y are 1 and 1 respectively.

Problem 3 :

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 ----(1)

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 (1) and (2) using elimination method.

x + y  =  548 ----(1)

2x + y  =  750 ----(2)

In the above two equations, y is having the same coefficient, that is 1.

Multiply the first equation by -1 to get the coefficient of -1. And keep the second equation as it is.

Then, we have

-x - y  =  - 548

2x + y  =  750

We can add the above two equations and eliminate y.

Then, we have

x  =  202

Step 4 :

Substitute 202 for x in the first equation.

(1)---->  202 + y  =  548

Subtract 202 from each side.

y  =  306

So, the number of adults tickets sold is 202 and the number of kids tickets sold is 346. Apart from the stuff given above, if you need any other stuff in math, please use our google custom search here.

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