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Word problems in algebra present a challenge to many students. The terminology and relation between different variables takes a bit of practice to understand. Algebra problem solver with steps free also explains how to go about writing those statements as equations. Once you put the equation into words, things become a easier and you just need to solve the equation. Practice sheets available with most online algebra problem solvers present students with the opportunity to get familiar with a wide variety of problems and scenarios.

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.
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:

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 = a2 + b- 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:

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:

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. answer
 

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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

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

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

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

= $\frac{6}{5}$

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

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

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

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 ax2 + bx + c = 0

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

b2 - 4ac = (- 45)2 - 4 * 1 * 1000

= 2025 - 4000

= - 1975

< 0

=> b2 - 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, $\frac{15a^2 - 10ab}{9a^2 - 4b^2}$

Step 1:

Solve numerator and denominator

15a2 - 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:

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


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

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