US Common Core K-12 Curriculum
Algebra : Solving Systems of Equations (Grade - 8)

Problem 1-8 : Solve each system of linear equations and write the solution as an ordered pair.

Problem 1 :

y = 3x - 3

2x + 5y = 19

Solution :

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

2x + 5y = 19 ----(2)

In (1), we have y = 3x - 3. So, plug in 3x - 3 for y in (2).

2x + 5(3x - 3) = 19

2x + 15x - 15 = 19

17x - 15 = 19

Add 15 to both sides of the equation.

17x = 34

Divide both sides by 17.

x = 2

Plug in x = 2 in (1).

y = 3(2) - 3

y = 6 - 3

y = 3

(x, y) = (2, 3)

Problem 2 :

y = 5 - 2x

4x + 3y = 13

Solution :

y = 5 - 2x ----(1)

4x + 3y = 13 ----(2)

In (1), we have y = 5 - 2x. So, plug in 5 - 2x for y in (2).

4x + 3(5 - 2x) = 13

4x + 15 - 6x = 13

15 - 2x = 13

Subtract 15 from both sides.

-2x = -2

Divide both sides by -2.

x = 1

Plug in x = 1 in (1).

y = 5 - 2(1)

y = 5 - 2

y = 3

(x, y) = (1, 3)

Problem 3 :

y = 3x + 2

y = 7x + 6

Solution :

y = 3x + 2 ----(1)

y = 7x + 6 ----(2)

Both the equations have been solved for y in terms of x. We can consider one of the two equations and take the stuff that we have for y.

Let's consider (2), in which, we have y = 7x + 6.

So, plug in 7x + 6 for y in (1).

7x + 6 = 3x + 2

Subtract 3x and 6 from both sides.

4x = -4

Divide both sides by 4.

x = -1

Now, we can use either (1) or (2) to solve for y.

Plug in x = -1 in either (1).

y = 3(-1) + 2

y = -3 + 2

y = -1

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

Problem 4 :

y = x + 3

y = x - 2

Solution :

y = x + 3 ----(1)

y = x - 2 ----(2)

Both the equations have been solved for y in terms of x. We can consider one of the two equations and take the stuff that we have for y.

Let's consider (1), in which, we have y = x + 3.

So, plug in x + 3 for y in (2).

x + 3 = x - 2

Subtract x from both sides.

3 = -2

In the above step, there is no x or y and 3 = 2 is false.

So, the given system has NO solution.

Problem 5 :

-2x + y = -5

-x - y = -7

Solution :

-2x + y = -5 ----(1)

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

Let us add (1) and (2) to eleiminate y.

(1) + (2) :

-3x = -12

Divide both sides by -3.

x = 4

Plug in x = 4 in (2).

-4 - y = -7

Add 4 to both sides.

-y = -3

Multiply both sides by -1.

y = 3

(x, y) = (4, 3)

Problem 6 :

2x + 5y = 19

x - 2y = -4

Solution :

2x + 5y = 19 ----(1)

x - 2y = -4 ----(2)

Let us multiply (2) by 2 and subtract it from (1) to eliminate x.

(1) - 2(2) :

9y = 27

Divide both sides by 9.

y = 3

Plug in y = 3 in (2).

x - 2(3) = -4

x - 6 = -4

Add 6 to both sides.

x = 2

(x, y) = (2, 3)

Problem 7 :

4x + 2y = 14

3x - 5y = -22

Solution :

4x + 2y = 14 ----(1)

3x - 5y = -22 ----(2)

Let us multiply (1) by 5, (2) by 2 and them to eliminate y.

5(1) + 2(2) :

26x = 26

Divide both sides by 26.

x = 1

Plug in x = 1 in (1).

4(1) + 2y = 14

4 + 2y = 14

Subtract 4 from both sides.

2y = 10

Divide both sides by 2.

y = 5

(x, y) = (1, 5)

Problem 8 :

2x - y = 4

6x - 3y = 12

Solution :

2x - y = 4 ----(1)

6x - 3y = 12 ----(2)

Let us multiply (1) by -3 and add it to (2) to eliminate y.

-3(1) + (2) :

0 = 0

In the above step, there is no x or y and 0 = 0 is true.

So, the given equation has infinitely many solutions.

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