13 Solve: 5 = 3. X - 3 X = +

Algebra: Structure And Method, Book 1
(REV)00th Edition
ISBN:9780395977224
Author:Richard G. Brown, Mary P. Dolciani, Robert H. Sorgenfrey, William L. Cole
Publisher:Richard G. Brown, Mary P. Dolciani, Robert H. Sorgenfrey, William L. Cole
Chapter3: Solving Equation And Problems
Section3.3: Using Several Transformations
Problem 19WE
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Solve:
+
2.
5
X + 2
This equation contains a rational expression that has a variable in the denominator. We begin by asking, "What value(s) of x make that denominator 0?"
If a number makes the denominator of a rational expression 0, that number cannot be a solution of the equation because division by 0 is undefined.
7
We note that x cannot be -2, because this would produce a 0 in the denominator of
х+ 2
Since the denominators of the rational expressions in the equation are 5 and x + 2, we multiply both sides by the LCD, 5(x + 2), to clear the equation of fractions.
3
+x+ 2
This is the equation to solve.
= 2
5
7
5(x + 2) (
E+z) = 5(x + 2)(2)
Write each side of the equation within parentheses and then multiply both sides by the LCD.
X + 2
sK* + 2(}) + s* + 2(,2)
7
5(x + 2)() + 5(x + 2)(-
5(x + 2)(2)
On the left side, distribute the multiplication by 5(x + 2).
7
1
X + 2
(x + 2)(5
+ 5
(x+2}
5(x + 2)(2)
On the left side, simplify:
1 and
X + 2
(X+2}
1.
1
3(x + 2) + 5(7)
(x + 2)
Simplify each side.
The resulting equation does not contain any fractions. We now solve this linear equation for x.
3x + 6 + 35 = 10x +
Use the distributive property and simplify.
Зх +
10x + 20
Combine like terms.
%3D
-7x =
Subtract 10x and 41 from both sides.
X =
Divide both sides by –7.
The solution is 3 and the solution set is {3}. To check, we substitute 3 for x in the original equation and simplify.
7
Check:
+
This is the original equation.
5
x + 2
7
+
Substitute.
5
3 + 2
3
7
5
2 =
True.
+
Transcribed Image Text:Solve: + 2. 5 X + 2 This equation contains a rational expression that has a variable in the denominator. We begin by asking, "What value(s) of x make that denominator 0?" If a number makes the denominator of a rational expression 0, that number cannot be a solution of the equation because division by 0 is undefined. 7 We note that x cannot be -2, because this would produce a 0 in the denominator of х+ 2 Since the denominators of the rational expressions in the equation are 5 and x + 2, we multiply both sides by the LCD, 5(x + 2), to clear the equation of fractions. 3 +x+ 2 This is the equation to solve. = 2 5 7 5(x + 2) ( E+z) = 5(x + 2)(2) Write each side of the equation within parentheses and then multiply both sides by the LCD. X + 2 sK* + 2(}) + s* + 2(,2) 7 5(x + 2)() + 5(x + 2)(- 5(x + 2)(2) On the left side, distribute the multiplication by 5(x + 2). 7 1 X + 2 (x + 2)(5 + 5 (x+2} 5(x + 2)(2) On the left side, simplify: 1 and X + 2 (X+2} 1. 1 3(x + 2) + 5(7) (x + 2) Simplify each side. The resulting equation does not contain any fractions. We now solve this linear equation for x. 3x + 6 + 35 = 10x + Use the distributive property and simplify. Зх + 10x + 20 Combine like terms. %3D -7x = Subtract 10x and 41 from both sides. X = Divide both sides by –7. The solution is 3 and the solution set is {3}. To check, we substitute 3 for x in the original equation and simplify. 7 Check: + This is the original equation. 5 x + 2 7 + Substitute. 5 3 + 2 3 7 5 2 = True. +
음 +
Solve:
X - 3
X =
3.
II
Transcribed Image Text:음 + Solve: X - 3 X = 3. II
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