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## Math Problem Solver

## Solved Examples

**Question 1:**Three years ago, father's age was the square of his son's age. Twelve years hence, his age will be twice that of his son's age. Find the present age of the father.

**Solution:**

Let the age of the son

Therefore age of his father = $x^2$

Therefore age of his father = $x^2$ + 15

The problem states :

Age of father = 2(age of his son)

=> $x^2$ + 15 = 2(x + 15)

=> x

=> x

=> x

=> x

x

=> x

=> x(x - 5) + 3(x - 5) = 0

=> (x + 3)(x - 5) = 0

either x + 3 = 0 or x - 5 = 0

=> x = -3 (neglecting negative value)

or x = 5

=> x = 5, son's age

Therefore father's age 3 years ago = 5

So father's present age = 25 + 3 = 28

Thus, the present age of the father = 28 years.

**3 years ago**= x yearsTherefore age of his father = $x^2$

Step 1:Step 1:

**Twelve years hence:****Age of Son = x + 15**

Therefore age of his father = $x^2$ + 15

Step 2:Step 2:

The problem states :

Age of father = 2(age of his son)

=> $x^2$ + 15 = 2(x + 15)

=> x

^{2}+ 15 = 2x + 30=> x

^{2}- 2x = 30 - 15=> x

^{2}- 2x = 15=> x

^{2}- 2x - 15 = 0**Step 3:**

Solve for x,x

^{2}- 2x - 15 = 0=> x

^{2}- 5x + 3x - 15 = 0=> x(x - 5) + 3(x - 5) = 0

=> (x + 3)(x - 5) = 0

Step 4:Step 4:

either x + 3 = 0 or x - 5 = 0

=> x = -3 (neglecting negative value)

or x = 5

=> x = 5, son's age

**3 years ago**Therefore father's age 3 years ago = 5

^{2}= 25So father's present age = 25 + 3 = 28

Thus, the present age of the father = 28 years.

**Question 2:**Solve the system of linear equations

6x + 15y = 50 and 3x + 10y = 50

**Solution:**

**Step 1:**

Given system of linear equations

6x + 15y = 50 ................................(1)

and

3x + 10y = 50 ..................................(2)

Step 2:

Step 2:

Multiply equation (2) by 2

=> 2(3x + 10y) = 2 * 50

=> 6x + 20y = 100 ...................................(3)

Step 3:

Step 3:

Solve (1) and (3):

Subtract equation( 3) from equation ( 1)

=> 6x + 15y - 6x - 20y = 50 - 100

=> - 5y = - 50

Divide each side by -5

=> y = 10

Step 4:

Step 4:

Put y = 10 in (1)

=> 6x + 15 * 10 = 50

=> 6x + 150 = 50

=> 6x = 50 - 150

=> 6x = - 100

Divide each side by 6, to isolate x,

=> x = $\frac{-100}{6}$

=> x = $\frac{-50}{3}$

Hence, the solution to the system, (x, y) = ($\frac{-50}{3}$, 10).