# Linear Equation Problems

Value of the unknown quantity for which from given equation we get true numerical equality is called root of that equation. Two equations are called equivalent when the multitudes of their roots match, the roots of the first equation are also roots of the second and vice versa. The following rules are valid:

1. If in given equation one expression is substituted with another identity one, we get equation equivalent to the given.

2. If in given equation some expression is transferred from one side to the other with contrary sign, we get equation equivalent to the given.

3. If we multiply or divide both sides of given equation with the same number, different from zero, we get equation equivalent to the given.

Equation of the kind $ax + b = 0$, where $a, b$ are given numbers is called
**simple equation** in reference to the unknown quantity $x$.

**Problem 1** Solve the equation:

A) 16x + 10 – 32 = 35 – 10x - 5

B) $y + \frac{3}{2y} + 25 = \frac{1}{2}y + \frac{3}{4}y – \frac{5}{2}y + y + 37$

C) 7u – 9 – 3u + 5 = 11u – 6 – 4u

**Solution**:

A)We do some of the makred actions and we get

16x – 22 = 30 – 10x

After using rule 2 we find 16x + 10x = 30 + 22

After doing the addition 26x = 52

We find unknown multiplier by dividing the product by the other multiplier.

That is why $x = \frac{52}{26}$

Therefore x = 2

B) By analogy with A) we find:

$y\left(1 + \frac{3}{2}\right) + 25 = y\left(\frac{1}{2} + \frac{3}{4} – \frac{5}{2} + 1\right) + 37 \Leftrightarrow$

$\frac{5}{2}y + 25 = -\frac{1}{4}y + 37 \Leftrightarrow \frac{5}{2}y + \frac{1}{4}y = 37 - 25 \Leftrightarrow$

$\frac{11}{4}y = 12 \Leftrightarrow y = \frac{12.4}{11} \Leftrightarrow y = \frac{48}{11}$

C) 4u – 4 = 7u – 6 <=> 6 – 4 = 7u – 4u <=> 2 = 3u <=> $u = \frac{2}{3}$

**Problem 2** Solve the equation :

A) 7(3x – 6) + 5(x - 3) - 2(x - 7) = 5

B) (x -3)(x + 4) - 2(3x - 2) = (x - 4)^{2}

C) (x + 1)^{3} – (x - 1)^{3} = 6(x^{2} + x + 1)

**Solution**:

A) 21x - 42 + 5x - 15 - 2x + 14 = 5<=>

21x + 5x - 2x = 5 + 42 + 15 - 14<=>

24x = 48 <=> x = 2

B) x^{2} + 4x - 3x - 12 - 6x + 4 = x^{2} - 8x + 16 <=>

x^{2} - 5x – x^{2} + 8x = 16 + 12 – 4 <=>

3x = 24 <=> x = 8

C) x^{3} + 3x^{2} + 3x + 1 – (x^{3} - 3x^{2} + 3x - 1) = 6x^{2} + 6x + 6 <=>

x^{3} + 3x^{2} + 3x + 1 – x^{3} + 3x^{2} + 1 = 6x^{2} + 6x + 6 <=>

2 = 6x + 6 <=> 6x = -4 <=> $x = -\frac{2}{3}$

**Problem 3** Solve the equation :

A) $\frac{5x - 4}{2} = \frac{0.5x + 1}{3}$

B) $1 –\frac{x - 3}{5} = \frac{-3x + 3}{3}$

C) $\frac{x + 1}{3} – \frac{2x + 5}{2} = -3$

D) $\frac{3(x - 1)}{2} + \frac{2(x + 2)}{4} = \frac{3x + 4.5}{5}$

**Solution**:

A) $\frac{5x - 4}{2} – \frac{0.5x + 1}{3} \Leftrightarrow$

3(5x - 4) = 2(0.5x + 1) <=>

15x - 12 = x + 2 <=>

15x – x = 12 + 2<=>

14x = 14 <=> x = 1

B) $1 – \frac{x - 3}{5} = \frac{3(1 - x)}{3}\Leftrightarrow$

$1 –\frac{x - 3}{5} = 1 – x\Leftrightarrow$

-x + 3 = - 5x <=>

5x – x = - 3 <=>

$x = -\frac{3}{4}$

C) $\frac{3(x - 1)}{2} + \frac{2(x + 2)}{4} = \frac{3x + 4.5}{5} \Leftrightarrow$

$\frac{2(x + 1) - 3(2x + 5)}{6} = - 3 \Leftrightarrow$

$\frac{2x + 2 - 6x -15}{6} = - 3 \Leftrightarrow$

-4x - 13 = -18 <=>

-4x = -18 + 13 <=>

-4x = -5 <=> $x = \frac{5}{4}$

D) We reduce to common denominator, which for 2, 4 and 5 is 20

$\frac{3(x - 1)}{2} + \frac{2(x + 2)}{4} = \frac{3x + 4.5}{5} \Leftrightarrow$

30(x - 1) + 10(x + 2) = 4(3x + 4.5) <=>

30x - 30 + 10x + 20 = 12x + 18 <=>

40x - 12x = 18 + 10 <=>

28x = 28 <=> x = 1

**Problem 4** Proof that every value of the unknown quantity is root of the equation:

A) 7x - 13 = - 13 + 7x

B) $(\frac{1}{2} – x)^2 – (\frac{1}{2} + x)^2 = -2x$

C) 3x - 3x = 26 - 2(7 + 6)

D) $\frac{-3x + 4x^2}{5} = (0.8x - 0.6)x$

**Solution**: For one simple equation with unknown quantity x everyx is a solution, if it is reduced to the following equivalent equation 0.x = 0 or it transforms into identity a = a. Actually, in the left any value of x, when we multiply it with zero, will obtain zero, i.e. the right side or the value of x won’t influence the right or left side of the identity.

A) 7x - 7x = -13 + 13 <=> 0.x = 0 <=> every x is a solution.

B) $\frac{1}{4} - x + x^2 –(\frac{1}{4} + x + x^2) = - 2x$ <=>

$\frac{1}{4} - x + x^2 -\frac{1}{4} – x – x^2 = - 2x$<=>

-2x = -2x <=>

-2x + 2x = 0 <=>

0.x = 0 <=> Therefore every x is a solution.

C) 0.x = 26 - 2.13 <=>

0.x = 26 – 26 <=>

0.x = 0 <=> every x is a solution.

D) -3x + 4x^{2} = 5(0.8x - 0.6)x <=>

-3x + 4x^{2} = (4x - 3)x <=>

-3x + 4x^{2} = 4x^{2} - 3x

Therefore every x is a solution.

**Problem 5** Proof that the equation has no roots:

A) 0.x = 34

B) 5 - 3x = 7 - 3x

C) $\frac{x - 3}{4} = \frac{x + 5}{4}$

D) 2(3x - 1) – 3(2x + 1) = 6

**Solution**:

A) For the left side we will get value 0 for every x and for the right side is 34, i.e. number different from 0. Therefore there is no such x to get a true numerical equality;

B) 5 - 3x = 7 - 3x <=> 3x - 3x = 7 - 5 <=> 0.x = 2 <=> 0 = 2, which is impossible for any x

C) $\frac{x - 3}{4} = \frac{x + 5}{4}$ <=> x - 3 = x + 5 <=> x – x = 5 + 3 <=> 0 = 8 => no solution;

D) 2(3x - 1) - 3(2x + 1) = 6 <=> 6x - 2 - 6x - 3 = 6 <=> 0.x = 6 + 5 <=> 0 = 11 no solution.

**Problem 6** Solve the equation:

A) 2x^{2} - 3(1 – x)(x + 2) + (x - 4)(1 - 5x) + 58 = 0

B) 3.(x + 1)^{2} – (3x + 5).x = x + 3

C) x^{2} – (x - 1).(x + 1) = 4

D) (x - 1).(x^{2} + x + 1) = (x - 1)^{3} + 3x(x - 1)

E) (3x - 1)^{2} – x(15x + 7) = x(x + 1).(x - 1) – (x + 2)^{3}

**Solution**:

A) 2x^{2} - 3(x + 2 – x^{2} - 2x) + x - 5x^{2} - 4 + 20x + 58 = 0 <=>

2x^{2} - 3x - 6 + 3x^{2} + 6x + x - 5x^{2} - 4 + 20x + 58 = 0 <=>

0.x^{2} + 24x + 48 = 0 <=>

24x = - 48 <=> x = -2

B) 3(x^{2} + 2x + 1) - 3x^{2} - 5x = 3x^{2} + 6x + 3 - 3x^{2} -5x = x + 3 <=>

(3 - 3)x^{2} + (6 - 5).x – x = 3 - 3 <=>

0 = 0 => every x is a solution

C) x^{2} – (x^{2} -1) = 4 <=>

x^{2} – x^{2} + 1 = 4 <=>

0 = 3 => no solution

D) x^{3} + x^{2} + x – x^{2} – x - 1 = x^{3} - 3x^{2} + 3x - 1 + 3x^{2} - 3x <=>

0 = 0 => every x is a solution

E) 9x^{2} - 6x + 1 - 15x^{2} - 7x = x^{3} –x^{2} + x^{2} – x – x^{3} - 6x^{2} - 12x - 8 <=>

0 = 9 => no solution

**Problem 7** Solve the equation:

A) $\frac{6x - 1}{5} - \frac{1 - 2x}{2} = \frac{12x + 49}{10}$

B) $\frac{x - 3}{2} + \frac{2x - 2}{4} = \frac{7x - 6}{3}$

**Solution**:

A)We reduce to common denominator and we get:

12x - 2 - 5 +10x = 12x + 49 <=>

22x - 12x = 49 + 7 <=>

10x = 56 <=> x = 5.6

B) $\frac{x - 3}{2} + \frac{2x - 2}{4} = \frac{7x - 6}{3}$ <=>

$\frac{x -3 + x - 1}{2} = \frac{7x - 6}{3}$ <=>

3(2x - 4) = 2(7x - 6) <=>

6x -12 = 14x - 12 <=>

8x = 0 <=> x = 0

**Problem 8** The function f(x) = x + 4 is given. Solve the equation:

$\frac{3f(x - 2)}{f(0)} + 4 = f(2x + 1)$

**Solution**:

We calculate f(0), f(x -2), f(2x +1), namely f(0) = 0 + 4 = 4;

f(x - 2) = x - 2 + 4 = x + 2;

f(2x + 1) = 2x + 1 + 4 = 2x + 5 The equation gets this look

$\frac{3(x + 2)}{4} + 4 = 2x + 5$ <=>

3(x + 2) +16 = 4(2x + 5) <=>

3x + 6 +16 = 8x + 20 <=>

22 - 20 = 8x - 3x <=>

2 = 5x <=> x = 0.4

**Problem 9** Solve the equation:

(2x - 1)^{2} – x(10x + 1) = x(1 – x)(1 + x) – (2 – x)^{3}

**Solution**:

(2x - 1)^{2} – x(10x + 1) = x(1 – x)(1 + x) – (2 – x)^{3} <=>

4x^{2} - 4x + 1 -10x^{2} – x = x – x^{3} - 8 + 12x - 6x^{2} + x^{3} <=>

18x = 9 <=> $x = \frac{1}{2}$

**Problem 10** Solve the equation:

(2x + 3)^{2} –x(1 + 2x)(1 - 2x) = (2x - 1)^{2} + 4x^{3} - 1

**Solution**:

(2x + 3)^{2} – x(1 + 2x)(1 - 2x) = (2x - 1)^{2} + 4x^{3} -1 <=>

4x^{2} + 12x + 9 – x(1 - 4x^{2}) = 4x^{2} - 4x + 1 + 4x^{3} - 1 <=>

12x + 9 – x + 4x^{3} = - 4x + 4x^{3}<=>

15x = -9 <=> $x = -\frac{3}{5}$

**Problem 11** Solve the equation :

(2x - 1)^{3} + 2x(2x - 3).(3 - 2x) – (3x - 1)^{2} = 3x^{2} - 2

**Solution**:

We open the brackets by using the formulas for multiplication:

8x^{3} - 3(2x)^{2}.1 + 3.2x(1)^{2} – 1^{3} - 2x(2x - 3)^{2} – (9x^{2} - 6x + 1) = 3x^{2} - 2 <=>

8x^{3} - 12x^{2} + 6x - 1 - 2x(4x^{2} - 12x + 9) - 9x^{2} + 6x - 1 = 3x^{2} - 2 <=>

8x^{3} - 21x^{2} + 12x - 8x^{3} + 24x^{2} - 9x = 3x^{2} <=>

3x^{2} + 3x = 3x^{2} <=>

3x = 0 <=> x = 0

**Problem 12** Solve the equation :

$\left(2x - \frac{1}{2}\right)^2 – (2x - 3)(2x + 3) = x + \frac{1}{4}$

**Solution**:

We use the formulas for multiplication, open the brackets and get:

$4x^2 - 2x + \frac{1}{4} – (4x - 9) = x + \frac{1}{4}$ <=>

$4x^2 - 2x + \frac{1}{4} - 4x^2 + 9 = x + \frac{1}{4}$ <=>

9 = x + 2x <=>

9 = 3x <=> x = 3

**Problem 13** Proof that the two equations are equivalent:

A) $\frac{x - 5}{2} + \frac{x - 1}{8} = \frac{1.5x - 10}{4}$ and $\frac{x + 6}{2} – \frac{5.5 - 0.5x}{3} = 1.5$

B) $x – \frac{8x + 7}{6} + \frac{x}{3} = -1\left(\frac{1}{6}\right)$ and $2x – \frac{6 – x}{3} - 2\left(\frac{1}{3}\right)x = -2$

**Solution**:

A) For the first equation we get consecutively:

4(x - 5) + x - 1 = 2(1.5x - 10) <=>

4x - 20 + x - 1 = 3x - 20 <=>

5x – 3x = - 20 + 21 <=>

2x = 1 <=> $x = \frac{1}{2}$,

for the second equation we have

3(x + 6) - 2(5.5 - 0.5y) = 6 . 1.5 <=>

3x + 18 - 11 + x = 9 <=>

4y = 9 - 7 <=>

$x = \frac{2}{4}$ <=> $x = \frac{1}{2}$ Therefore the equations are equivalent.

B) Analogical to A) Try by yourself

**Problem 14**Solve the equation :

A) (2x + 1)^{2} – x(1 - 2x)(1 + 2x) = (2x - 1)^{2} + 4x^{3} - 3

B) (2x - 1)^{2} + (x - 2)^{3} = x^{2}(x - 2) + 8x - 7

C) (x + 2)(x^{2} - 2x + 4) + x(1 – x)(1 + x) = x - 4

D) $\frac{8x + 5}{4} – \frac{1}{2\left[2 – \frac{3 – x}{3}\right]} = 2x + \frac{5}{6}$

E) $\frac{x}{3} – \frac{x + 3}{4} = x – \frac{1}{3\left[1 – \frac{3 - 24x}{8}\right]}$

F) $\frac{x}{5} – \frac{(2x - 3)^2}{3} = \frac{1}{5}\left[5 – \frac{20x - 43x}{3}\right]$

**Solution**:

A) 4x^{2} + 4x + 1 – x(1 - 4x^{2}) = 4x^{2} - 4x + 1 + 4x^{3} - 3 <=>

4x – x + 4x^{3} = -4x + 4x^{3} -3 <=>

3x + 4x = -3 <=>

7x = - 3 <=> $x = -\frac{3}{7}$

B) 4x^{2} - 4x + 1 + x^{3} - 3x^{2}.2 + 3x.2^{2} - 8 = x^{3} -2x^{2} + 8x - 7 <=>

4x^{2} - 6x^{2} - 4x + 1 + 12x - 8 = - 2x^{2} + 8x -7 <=>

-2x^{2} + 8x - 7 = - 2x^{2} + 8x - 7 <=>

0 = 0 => every x is a solution;

C) x^{3} + 2x^{2} - 2x^{2} - 4x + 4x + 8 + x(1 – x^{2}) = x - 4 <=>

x^{3} + 8 + x – x^{3} = x - 4 <=>

8 = - 4, which is impossible. Therefore the equation has no solution;

D) $\frac{8x + 5}{4} - 1 + \frac{3 – x}{6} = 2x + \frac{5}{6} \Leftrightarrow$

3(8x + 5) - 12 + 2(3 – x) = 24x + 2.5 <=>

24x + 15 - 12 + 6 - 2x = 24x + 10 <=>

-2x = 10 - 9 <=> $x = -\frac{1}{2}$

E) $\frac{x}{3} – \frac{x + 3}{4} = x - \frac{1}{3} + \frac{3 - 24x}{24} \Leftrightarrow$

8x - 6(x + 3) = 24x - 8 + 3 - 24x <=>

8x - 6x - 18 = -5 <=>

2x = 18 - 5 <=>

2x = 13 <=> x = 6.5

F) $\frac{x}{5} – \left[\frac{2x - 3}{3}\right]^2 = 1 – \frac{20x^2 - 43x}{15} \Leftrightarrow$

3x - 5(4x^{2} -12x + 9) = 15 - 20x^{2} + 43x <=>

3x - 20x^{2} + 60x - 45 = 15 - 20x^{2} + 43x <=>

63x - 43x = 15 + 45 <=>

20x = 60 <=> x = 3

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