Solve the system of equations using elimination: minus, 3, x, plus, 2, y, equals, 10−3x+2y=10 and 5, x, minus, 5, y, equals, minus, 105x−5y=−10

Answer 1

To solve the system of equations using elimination, we will eliminate one of the variables by adding or subtracting the equations. Let's eliminate the x variable.

First, let's multiply the second equation by 3 to make the x coefficients equal:

(3) × (5x - 5y = -10)
15x - 15y = -30

Now, we can add the new second equation to the first equation:

(10 - 3x + 2y) + (15x - 15y) = 10

Combine like terms:

10 + 15x - 3x + 2y - 15y = 10

Simplify:

12x - 13y = 0

Now, we have a new equation:

12x - 13y = 0 .............(3)

Let's use equations (1) and (3) to eliminate the x variable. Multiply equation (1) by 12:

12(10 - 3x + 2y) = 12(10)

120 - 36x + 24y = 120

Now, we can subtract equation (3) from the new equation:

(120 - 36x + 24y) - (12x - 13y) = 120

120 - 36x + 24y - 12x + 13y = 120

Combine like terms:

-48x + 37y = 0

Now, we have another new equation:

-48x + 37y = 0 .............(4)

Now, we have a system of equations:

12x - 13y = 0 .............(3)
-48x + 37y = 0 .............(4)

To eliminate the x variable, let's multiply equation (4) by 3:

3(-48x + 37y) = 3(0)

-144x + 111y = 0

Now, we can subtract equation (3) from equation (4):

(-144x + 111y) - (12x - 13y) = 0

-144x + 111y - 12x + 13y = 0

Combine like terms:

-156x + 124y = 0

Now, we have a new equation:

-156x + 124y = 0 .............(5)

Equation (5) is equivalent to equation (4).

Therefore, the system of equations is:

12x - 13y = 0 .............(3)
-156x + 124y = 0 .............(5)

To solve the system, we can set one of the variables to a constant, such as y = 1.

Let's solve equation (3) for x:

12x - 13y = 0
12x - 13(1) = 0
12x - 13 = 0
12x = 13
x = 13/12

So, the solution to the system of equations is x = 13/12 and y = 1.