Solving Equations with One Variable Worksheet

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 (- ∞,∞).

Correct answer : (2)

Which of the following are the solutions of the equation

\frac{3 x}{x + 5} + \frac{1}{x + 2}

\frac{7}{x^{2} + 7 x + 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)
[Use quadratic formula.]

= -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

Correct answer : (1)

Solve the equation

\frac{x + 5}{x} - \frac{7}{x + 2} - \frac{26}{x^{2} + 2 x}

= 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.]

Correct answer : (1)

Solve the equation

\frac{x + 8}{x} - \frac{2}{x + 2}

\frac{1}{x^{2} + 2 x}

	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.

Correct answer : (3)

Solve the equation

\frac{x - 10}{11}

\frac{x + 12}{11}

\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 .

Correct answer : (2)

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

Correct answer : (4)

Which of the following are the solutions of the equation

x

\frac{9}{x - 10}

= 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)
[Use Quadratic formula.]

= 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.]

Correct answer : (3)

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.

Correct answer : (1)

Solve :

\frac{x^{2}}{x - 7}

< 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 , ∞).

Correct answer : (4)

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, ∞).

Correct answer : (4)