# SOLVING SYSTEMS BY ELIMINATION WORKSHEET

## About "Solving systems by elimination worksheet"

Solving systems by elimination worksheet :

Worksheet given in this section is much useful to the students who would like to practice problems on solving system of linear equations in two variables.

## Solving systems by elimination : Steps

Step 1 :

The variable which is eliminated must have the same coefficient in both the equations. If not, make them to be same using LCM and multiplication.

Step 2 :

The variable which is eliminated must have different signs. If not, multiply one of the equations by negative sign.

Step 3 :

Now add the two equations to eliminate the variable.

## Solving systems by elimination worksheet - Problems

1. Solve the system of equations using elimination method. Check the solution by graphing.

2x - 3y  =  12

x + 3y  =  6

2.  Sum of the cost price of two products is \$50. Sum of the selling price of the same two products is \$52. If one is sold at 20% profit and other one is sold at 20% loss, find the cost price of each product.

## Solving systems by elimination worksheet - Solution

Problem 1 :

Solve the system of equations using elimination method. Check the solution by graphing.

2x - 3y  =  12

x + 3y  =  6

Solution :

Step 1 :

In the given two equations, the variable y is having the same coefficient (3). And also, the variable y is having different signs.

So we can eliminate the variable y by adding the two equations.

Divide both sides by 3.

3x / 3  =  18 / 3

x  =  6

Step 2 :

Plug x  =  6 in one of the equations.

x + 3y  =  6

6 + 3y  =  6

Subtract 6 from both sides.

aaaaaaaaaaaaaaaaaaaaaa 6 + 3y  =  6 aaaaaaaaaaaaaaaaaaa aaaaaaaaaaaaaaaaaaaa  - 6           - 6 aaaaaaaaaaaaaaaaaaa aaaaaaaaaaaaaaaaaaaa  -------------- aaaaaaaaaaaaaaaaaaa aaaaaaaaaaaaaaaaaaaa          3y  =  0 aaaaaaaaaaaaaaaaaaa aaaaaaaaaaaaaaaaaaaa  -------------- aaaaaaaaaaaaaaaaaaa

Divide both sides by 3

3y / 3  =  0 / 3

y  =  0

Step 3 :

Write the solution as ordered pair as (x, y).

(6, 0)

Step 4 :

Check the solution by graphing.

To graph the equations, write them in slope-intercept form.

That is,

y  =  mx + b

2x - 3y  =  12

y  =  (2/3)x - 4

Slope  =  2/3

y-intercept  =  -4

x + 3y  =  6

y  =  -(1/3)x + 2

Slope  =  -1/3

y-intercept  =  2

The point of intersection is (6, 0).

Problem 2 :

Sum of the cost price of two products is \$50. Sum of the selling price of the same two products is \$52. If one is sold at 20% profit and other one is sold at 20% loss, find the cost price of each product.

Solution :

Step 1 :

Let "x" and "y" be the cost prices of two products.

Then,  x + y  =  50  --------(1)

Step 2 :

Let us assume that "x" is sold at 20% profit

Then, the selling price of "x" is 120% of "x"

Selling price of "x"  =  1.2x

Let us assume that "y" is sold at 20% loss

Then, the selling price of "y" is 80% of "y"

Selling price of "x"  =  0.8y

Given : Selling price of "x"  +  Selling price of "y"  =  52

1.2x + 0.8y  =  52

To avoid decimal, multiply both sides by 10

12x + 8y  =  520

Divide both sides by 4.

3x + 2y  =  130 --------(2)

Step 3 :

Eliminate one of the variables to get the value of the other variable.

In (1) and (2), both the variables "x" and "y" are not having the same coefficient.

One of the variables must have the same coefficient.

So multiply both sides of (1) by 2 to make the coefficients of "y" same in both the equations.

(1) ⋅ 2 -------->  2x + 2y  =  100 ----------(3)

Variable "y" is having the same sign in both (2) and (3).

To change the sign of "y" in (3), multiply both sides of (3) by negative sign.

- (2x + 2y)  =  - 100

- 2x - 2y  =  - 100 --------(4)

Step 4 :

Now, eliminate the variable "y"in (2) and (4) as given below and find the value of "x".

Step 5 :

Plug x  =  30 in (1) to get the value of y.

(2) --------> 30 + y  =  50

Subtract 30 from both sides.

aaaaaaaaaaaaaaaaaaaa    30 + y  =  50 aaaaaaaaaaaaaaaaa aaaaaaaaaaaaaaaaaaa   - 30         - 30 aaaaaaaaaaaaaaaaa aaaaaaaaaaaaaaaaaaa------------------- aaaaaaaaaaaaaaaaa aaaaaaaaaaaaaaaaaaa             y   =  20 aaaaaaaaaaaaaaa  aaaaaaaaaaaaaaaaaaa------------------- aaaaaaaaaaaaaaaaa

Hence, the cost prices of two products are \$30 and \$20.

After having gone through the stuff given above, we hope that the students would have understood "Solving systems by elimination worksheet".

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