﻿ Solving Equations with One Variable Worksheet | Problems & Solutions

# Solving Equations with One Variable Worksheet

Solving Equations with One Variable Worksheet
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1.
Solve : (6$x$2 + 9) (2$x$4 + 5) > 0
 a. (0 , ∞) only b. (- ∞ , ∞) c. (- ∞ , 0) only d. Integers only

#### Solution:

Let f(x) = (6x2 + 9) (2x4 + 5)

(6x2 + 9) is always positive for all real values of x.
[x2 > 0.]

(2x4 + 5) is always positive for all real values of x.
[x4 > 0.]

So, f(x) = (6x2 + 9) (2x4 + 5) is positive for all real values of x.

The solution of the inequality is (- ∞,∞).

2.
Which of the following are the solutions of the equation $\frac{3x}{x+5}+\frac{1}{x+2}$ = $\frac{7}{{x}^{2}+7x+10}$?
 a. $x$ = - 2.590 and 0.2573 b. $x$ = 2 c. $x$ = - 5.180 and 0.5146 d. $x$ = - 7.2081 and 0.2081

#### Solution:

The LCD of the equation 3xx+5+1x+2 = 7x2+7x+10 is x2 + 7 x +10.

3x (x + 2) + (x + 5) = 7
[Multiply both sides of the equation by LCD x2 + 7 x +10 of the equation.]

3 x2 + 6x + x + 5 = 7

3x2 + 7x + 5 = 7

3x2 + 7x - 2 = 0

x = -7 ±72 - 4(3)(-2)2(3)

= -7 ±49+246

= -7 ±736

= -7 ± 8.5446

x = - 2.590 and 0.2573
[Use a calculator.]

So, the solutions of the equation are -2.590 and 0.2573

3.
Solve the equation = 0.
 a. $x$ = 4 b. $x$ = ± 4 c. $x$ = 16 d. $x$ = ± 16

#### Solution:

The LCD of the equation x + 5x-7x + 2-26x2+2x = 0 is x2 + 2x.

(x + 5)(x + 2) - 7x - 26 = 0
[Multiply both sides of the equation by the LCD (x2 + 2x) of the equation.]

x2 + 7x + 10 - 7x - 26 = 0

x2 - 16 = 0

x = ± 4

So, the solution of the equation is x = 4.
[The equation is not defined for x = - 4.]

4.
Solve the equation $\frac{x+8}{x}-\frac{2}{x+2}$ = $\frac{1}{{x}^{2}+2x}$.
 a. $x$ = 5, - 3 b. $x$ = 5, 3 c. $x$ = - 5, - 3 d. $x$ = - 5, 3

#### Solution:

The LCD of the equation x + 8x-2x + 2 = 1x2 + 2x is x2 + 2x.

(x + 8)(x + 2) - 2x = 1
[Multiply both sides of the equation by the LCD x2 + 2x of the equation.]

x2+10x+16-2x = 1

x2 + 8x + 15 = 0

(x + 5)(x + 3) = 0
[Factor.]

x + 5 = 0 or x + 3 = 0

x = - 5 and - 3

So, the solutions of the given equation are - 5, - 3.

5.
Solve the equation + = $\frac{12}{11}$.
 a. $x$ = -5 b. $x$ = 5 c. $x$ = 10 d. $x$ = -10

#### Solution:

The LCD of the equation x - 1011 + x + 1211 = 1211 is 11.

x - 10 + x + 12 = 12
[Multiply both sides of the equation by the LCD 11 of the equation.]

2x + 2 = 12

2x = 10 and hence x = 5 .

6.
Which of the following are the solutions of the equation $x$ - $\frac{88}{x}$ = 3.
 a. $x$ = 0, 8 b. $x$ = - 11, 8 c. $x$ = - 8 , - 11 d. $x$ = 11 , - 8

#### Solution:

The LCD of the equation x - 88x = 3 is x

x2 - 88 = 3x
[Multiply both sides of the equation by the LCD x of the equation.]

x2 - 3x - 88 = 0
[Factor.]

(x - 11)(x + 8) = 0

x - 11 = 0 or x + 8 = 0

x = 11 and - 8

So, the solutions of the equation are - 8 and 11

7.
Which of the following are the solutions of the equation $x$ + = 0?
 a. $x$ = 2 and - 18 b. $x$ = 1 and - 9 c. $x$ = 1 and 9 d. $x$ = 2 and 18

#### Solution:

The LCD of the equation x + 9x - 10 = 0 is x - 10.

x (x - 10) + 9 = 0
[Multiply both sides of the eqation by the LCD x - 10 of the equation.]

x2 - 10x + 9 = 0

x = 10±(- 10)2 - 4(1)(9)2(1)

= 10±100 - 362

= 10±642

Hence, x = 10 +642, 10 -642

So, the solutions of the equation are x = 1 and 9.
[Use a calculator.]

8.
Sum of a number and its reciprocal is - 2. Find the number.
 a. - 1 b. - 2 c. - 1, - 2 d. $\frac{1}{2}$

#### Solution:

Let x be the number such that x + 1x = - 2.

x2 + 1 = - 2x
[Multiply the equation by the LCD of the equation x.]

x2 + 2x + 1 = 0

(x +  1)2 = 0

x +  1 = 0 and hence x = - 1.

So, the required number is - 1.

9.
Solve : < 0
 a. (- ∞ , 7) b. (- ∞ , ∞) c. [7 , ∞) d. (7 , ∞)

#### Solution:

Let f(x) = x2x - 7

Since x2 is always positive for all real values of x except 0, the sign of f(x) depends on the sign of the factor (x - 7).

If x - 7 < 0 or x < 7, then f(x) is negative.

If x - 7 > 0 or x > 7, then f(x) is positive.

So, the solution of the inequality is (7 , ∞).

10.
Solve ($x$ + 7)| $x$ - 5 | ≥ 0.
 a. (- 5, ∞] b. [- 5, ∞) c. (- 7, ∞] d. [- 7, ∞)

#### Solution:

Let f (x) = (x + 7)| x - 5 |

The factor |x - 5| is positive for all the values of x except 5, and is zero at x = 5.

The factor (x + 7) is positive for all the values of x > - 7 and is zero at x = - 7.

So, the solution of the given inequality is [- 7, ∞).