**Solving linear systems using multiplication and addition worksheet :**

In some linear systems, neither variable can be eliminated by adding or subtracting the equations directly. In systems like these, you need to multiply one of the equations by a constant so that adding or subtracting the equations will eliminate one variable.

1. Solve the system of equations by multiplying and adding.

3x - 5y = -17

2x + 15y = 7

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.

**Problem 1 :**

Solve the system of equations by multiplying and adding.

3x - 5y = -17

2x + 15y = 7

**Solution :**

**Step 1 :**

Let us eliminate the variable y in the given two equations.

3x - 5y = -17 -------- (1)

2x + 15y = 7 -------- (2)

**Step 2 :**

To make the coefficient of y same in both the equations, multiply the first equation by 3.

(1) **⋅** 3 ----- > 9x - 15y = -51 -------- (3)

In equations (2) and (3), the variable y is having the same coefficient, but having different signs.

**Step 3 :**

Add the equations (2) and (3) to eliminate the variable y.

Divide both sides by 11.

11x / 11 = - 44 / 11

x = - 4

**Step 4 : **

Substitute the value of x into one of the equations to find the value of y.

3x - 5y = -17

3(-4) - 5y = -17

-12 - 5y = -17

Add 12 to both sides.

(-12 - 5y) + 12 = (-17) + 12

-12 - 5y + 12 = -17 + 12

Simplify.

-5y = -5

Divide both sides by -5

-5y / (-5) = -5 / (-5)

y = 1

Hence, the solution to the system is

(x, y) = (-4, 1)

**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 : **

Add the equations (2) and (3) to eliminate the variable y.

**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 aaaaaaaaaaaaaaaaaa aaaaaaaaaaaaaaaaaaa------------------- aaaaaaaaaaaaaaaaa

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

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