Question

5x+3y=35 2x+4y=28 Solve by elimination, cross multiplication, substituition and reduction.

Answer 1

x = 4 and y = 5

Step-by-step explanation:

2x + 4y= 28 ⇔ 2x = 28 - 4y ⇔ x = 14 - 2y

5x + 3y = 35 ⇔ 5(14 - 2y) + 3y = 35 ⇔ 70 - 10y + 3y = 35

⇔ -7y + 70 = 35

⇔ 7y = 70 - 35 = 35

⇔ y = 35/7 = 5

2x + 4y= 28 ⇔ 2x + 20 = 28 ⇔ 2x = 8 ⇔ x = 8/2 = 4

Answer 2

The solution set for the system of equations is (x, y) = (4, 5).

Step-by-step explanation:

1. Elimination: Multiply the first equation by 2 and the second equation by 3 to allow for the elimination of one variable.

10x + 6y = 70

6x + 12y = 84

Subtract the first equation from the second:

-4x + 6y = 14

Solve for x:

x = (70 - 14) / 4 = 14

Substitute x = 4 into the first equation:

5*4 + 3y = 35

20 + 3y = 35

3y = 35 - 20 = 15

y = 15/3 = 5

2. Substitution: From 5x + 3y = 35, we can express x as x = (35 - 3y) / 5.

Substitute into the equation 2x + 4y = 28:

2((35 - 3y) / 5) + 4y = 28

Solve for y to get y = 5, then substitute y = 5 into x = (35 - 3y) / 5 to get x = 4.

3. Cross Multiplication or Cramer's Rule:

D = |5 3| = (5*4) - (2*3) = 14

Dx = |35 3| = (35*4) - (28*3) = 56

Dy = |5 35| = (5*28) - (35*2) = 70

x = Dx / D = 56 / 14 = 4

y = Dy / D = 70 / 14 = 5

4. Reduction: The two equations can be rewritten and divided:

(5x + 3y) / 35 = 1

(2x + 4y) / 28 = 1

Solving this system yields x = 4, y = 5.

In each method, the solution set (x, y) = (4, 5).