Question

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

Answer 1

Let us write the equations again.

5x + 3y = 35
2x + 4y = 28

To solve this, it is best to eliminate the equation with the least coefficient of the unknown. Since the least coefficient is 2, we will eliminate x.

So, multiply equation1 by 2 and equation2 by 5. On doing this, we get

10x + 6y = 70 and
10x + 20y = 140

Subtracting the two equations, we get

14y = 70 or y = 70/14 = 5.

Substituting the value of 5 in equation2, we get

2x + 20 = 28 or

2x = 28 - 20 = 8.

Thus, x = 8/2 = 4.

Thus, x = 4 and y = 5.

Answer 2

Step-by-step explanation:

ELIMINATION METHOD :- multiply 1st equation by 2 and 2nd equation by 5 We get

10x+6y=70

10x+20y=140 by solving we get

Y=5 and X=4

Now BY SUBSTITUTION METHOD:--

Put x =35-3y/5 from 1st equation in 2nd equation

we get

1/5(70-6y )+4y=28

by solving we get

X=4 and Y=5

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