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Solved Example

**Question:**Divide 16 into two equal parts such that twice the square of the greater part is 164 more than the square of the smaller part.

**Solution:**

Let two parts be x and y, and x is greater than y.

Divide 16 into two equal parts

=> x + y = 16 ..............................(1)

and twice the square of the greater part = 164 + square of the smaller part

=> 2$x^2$ = 164 + $y^2$ .................................(2)

Solve (1) and (2)

(1) => y = 16 - x, put in (2)

=> 2$x^2$ = 164 + $(16 - x)^2$

=> 2$x^2$ = 164 + $16^2$ + $x^2$ - 2 * 16 * x

[(a - b)

=> 2$x^2$ = 164 + 256 + $x^2$ - 32x

=> 2$x^2$ = 420 + $x^2$ - 32x

=> 2$x^2$ - $x^2$ + 32x - 420 = 0

=> $x^2$ + 32x - 420 = 0

Solve for x, $x^2$ + 32x - 420 = 0

=> $x^2$ + 42x - 10x - 420 = 0

=> x(x + 42) - 10(x + 42) = 0

=> (x - 10)(x + 42) = 0

either x - 10 = 0 or x + 42 = 0

=> x = 10 or x = - 42, neglected

So x = 10 is only the value of x.

(1) => 10 + y = 16

=> y = 16 - 10 = 6

Hence, the parts are: 10 and 6.

**Step 1:**Divide 16 into two equal parts

=> x + y = 16 ..............................(1)

and twice the square of the greater part = 164 + square of the smaller part

=> 2$x^2$ = 164 + $y^2$ .................................(2)

Step 2:Step 2:

Solve (1) and (2)

(1) => y = 16 - x, put in (2)

=> 2$x^2$ = 164 + $(16 - x)^2$

=> 2$x^2$ = 164 + $16^2$ + $x^2$ - 2 * 16 * x

[(a - b)

^{2}= a^{2}+ b^{2 }- 2ab]=> 2$x^2$ = 164 + 256 + $x^2$ - 32x

=> 2$x^2$ = 420 + $x^2$ - 32x

=> 2$x^2$ - $x^2$ + 32x - 420 = 0

=> $x^2$ + 32x - 420 = 0

Step 3:Step 3:

Solve for x, $x^2$ + 32x - 420 = 0

=> $x^2$ + 42x - 10x - 420 = 0

=> x(x + 42) - 10(x + 42) = 0

=> (x - 10)(x + 42) = 0

Step 4:Step 4:

either x - 10 = 0 or x + 42 = 0

=> x = 10 or x = - 42, neglected

So x = 10 is only the value of x.

(1) => 10 + y = 16

=> y = 16 - 10 = 6

Hence, the parts are: 10 and 6.

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## Solved Examples

**Question 1:**If x : y = $\frac{3}{2}$, Find the value of $\frac{4x + 6y}{2x + 7y}$

**Solution:**

Since x : y = 3 : 2

Let x = 3k and y = 2k

=> $\frac{4x + 6y}{2x + 7y}$ = $\frac{4 * 3k + 6 * 2k}{2 * 3k + 7 * 2k}$

= $\frac{12k + 12k}{6k + 14k}$

= $\frac{24k}{20k}$

= $\frac{6}{5}$

=> $\frac{4x + 6y}{2x + 7y}$ = $\frac{6}{5}$

Let x = 3k and y = 2k

=> $\frac{4x + 6y}{2x + 7y}$ = $\frac{4 * 3k + 6 * 2k}{2 * 3k + 7 * 2k}$

= $\frac{12k + 12k}{6k + 14k}$

= $\frac{24k}{20k}$

= $\frac{6}{5}$

=> $\frac{4x + 6y}{2x + 7y}$ = $\frac{6}{5}$

**Question 2:**Solve (x - 5)(40 - x) = 800 and find the real roots.

**Solution:**

Given (x - 5)(40 - x) = 800

(x - 5)(40 - x) = 800

=> x(40 - x) - 5(40 - x) = 800

=> 40x - $x^2$ - 200 + 5x = 800

=> 45x - $x^2$ - 1000 = 0

or $x^2$ - 45x + 1000 = 0

Which is quadratic equation

Solve for x, $x^2$ - 45x + 1000 = 0

=> $x^2$ - 45x + 1000 = 0

Compare with ax

a = 1, b = - 45 and c = 1000

b

= 2025 - 4000

= - 1975

< 0

=> b

Hence this quadratic equation doesn't have any real root.

Step 1:Step 1:

(x - 5)(40 - x) = 800

=> x(40 - x) - 5(40 - x) = 800

=> 40x - $x^2$ - 200 + 5x = 800

=> 45x - $x^2$ - 1000 = 0

or $x^2$ - 45x + 1000 = 0

Which is quadratic equation

**Step 2:**Solve for x, $x^2$ - 45x + 1000 = 0

=> $x^2$ - 45x + 1000 = 0

Compare with ax

^{2}+ bx + c = 0a = 1, b = - 45 and c = 1000

b

^{2}- 4ac = (- 45)^{2}- 4 * 1 * 1000= 2025 - 4000

= - 1975

< 0

=> b

^{2}- 4ac < 0 ( no real roots)Hence this quadratic equation doesn't have any real root.

**Question 3:**Write $\frac{15a^2 - 10ab}{9a^2 - 4b^2}$

**Solution:**

Given, $\frac{15a^2 - 10ab}{9a^2 - 4b^2}$

Solve numerator and denominator

15a

and

9$a^2$ - 4$b^2$ = $(3a)^2$ - $(2b)^2$

= (3a - 2b)(3a + 2b)

[ $a^2$ - $b^2$ = (a - b)(a + b)]

$\frac{15a^2 - 10ab}{9a^2 - 4b^2}$ = $\frac{5a(3a - 2b)}{(3a - 2b)(3a + 2b)}$

= $\frac{5a}{3a + 2b}$

=> $\frac{15a^2 - 10ab}{9a^2 - 4b^2}$ = $\frac{5a}{3a + 2b}$.

Step 1:Step 1:

Solve numerator and denominator

15a

^{2}- 10ab = 5a(3a - 2b)and

9$a^2$ - 4$b^2$ = $(3a)^2$ - $(2b)^2$

= (3a - 2b)(3a + 2b)

[ $a^2$ - $b^2$ = (a - b)(a + b)]

**Step 2**:$\frac{15a^2 - 10ab}{9a^2 - 4b^2}$ = $\frac{5a(3a - 2b)}{(3a - 2b)(3a + 2b)}$

= $\frac{5a}{3a + 2b}$

=> $\frac{15a^2 - 10ab}{9a^2 - 4b^2}$ = $\frac{5a}{3a + 2b}$.