Question

ten years ago the age of a father was four times of his son ten years hence the age of the father will be twice that of his son the present ages of the father and the son are only only​

Answer 1

Answer:

Let the present ages of the father and son be xx and yy respectively.

10 years ago:

We know that the age of the father was 4 times the age of his son. So, we can express this relationship in terms of xx and yy;

x−10=4(y−10)x−10=4(y−10)

x−10=4y−40x−10=4y−40

x=4y−30x=4y−30 ——(1)

This equation will come in handy later on.

10 years after;

Now we know that the age of the father will be 2 times the age of his son.

x+10=2(y+10)x+10=2(y+10)

x+10=2y+20x+10=2y+20

x−2y=20−10x−2y=20−10

x−2y=10x−2y=10 ——(2)

So, in the end we have two equations (1) and (2), in terms of xx and yy, so we have two simultaneous equations and we can solve this by using substitution.

Substitution Method:

The two equations are:

x=4y−30x=4y−30——(1)

x−2y=10x−2y=10 ——(2)

In the substitution method, one variable is expressed in terms of another. This would allow the number of variables to be reduced to one, and so we can solve for that one variable. Once we get the value of that variable, we use it to solve for the other variable. Let’s see how this works;

So, since we have x=4y−30x=4y−30, I will choose to substitute (“replace”) every instance of xx with 4y−304y−30, and so reducing the number of variables to just one, namely yy. We can substitute this in equation (2);

x−2y=10x−2y=10

(4y−30)−2y=10(4y−30)−2y=10

hope it helps you

please mark me as the brainliest