Linear Equations in One Variable
For \(a , b \in R \) and \( a \neq 0\), equalities in the form of \( ax +b = 0 \) are called linear equations in one variable.
The number \( x = \displaystyle{ -\frac{b}{a} } \) that satisfies the equality \( ax +b = 0 \) is called the root of this equation, and the solution set of the equation is given by:
Examples:
\( \bullet \quad \frac{2x – 1}{3} + \frac{3x+2}{6} = 2 \) Let’s find the solution set of the equation.
\[ \frac{2x \ – \ 1 }{3} + \frac{3x +2}{6} = 2 \]
\[ \Rightarrow \frac{2(2x \ – \ 1) }{2 \cdot 3} + \frac{3x+2}{6} = 2 \]
\[\Rightarrow \frac{4x \ – \ 2 + 3x+ 2 }{6} = 2 \]
\[ \Rightarrow 7x = 12 \]
\[ \Rightarrow x = \frac{12}{7} \]
\[ \Rightarrow S = \left\{\frac{12}{7} \right\} \]
Warning:
In rational equations of the form \[ \frac{ P(x)}{Q(x)} = 0 \] we must have \( P(x) =0 \) and \( Q(x) \neq 0 \). The values of \(x\) that make the denominator zero make the fraction undefined, and therefore they cannot be elements of the solution set.
In equations of the form \[ P(x) \cdot Q(x) = 0 \] we must have \( P(x) =0 \) or \( Q(x) =0 \).
Question 1
\[ \frac{x}{x-2} + \frac{1}{2x-1} = \frac{4}{x^2-4} + \frac{2x}{2x-1} \]
What is the solution set of the equation?
\[
\text{A)} \{ 0,2 \} \quad
\text{B) } \{ 2 \} \quad
\text{C) } \{ 0 \} \quad
\text{D) } \{ 1,2 \} \quad
\text{E) } \{ 3 \}
\]
Solution:
\[ \frac{x}{x \ – \ 2} + \frac{1}{2x \ – \ 1} = \frac{4}{x^2 \ – \ 4} + \frac{2x}{2x \ – \ 1} \]
\[ \Rightarrow \frac{x}{x-2} =\frac{4}{x^2 \ – \ 4} + \frac{2x \ – \ 1}{2x \ – \ 1} \]
\[ \Rightarrow \frac{x}{x-2} = \frac{4 + x^2 \ – \ 4}{x^2 \ – \ 4} \]
\[ \Rightarrow x(x^2-4) = (x \ – \ 2)x^2 \]
\[\Rightarrow 2x (x \ – \ 2) =0 \]
\[ \Rightarrow x = 0 \quad \text{or} \quad x= 2 \]
Here, for
\[ x = 2 \Rightarrow \frac{x}{x \ – \ 2} \quad \text{and} \quad \frac{4}{x^2 \ – \ 4} \]
the denominators of the fractions become 0, so the number 2 cannot be an element of the solution set. Therefore,
\[ S = \{ 0 \} \]
\(\textbf{Answer: C} \)
Question 2
\[ \displaystyle \frac{ \displaystyle \frac{x}{x \ – \ 1} + \displaystyle \frac{1}{1 \ – \ x} }{x} = \frac{x}{2 \ – \ x} \]
What is the solution set of the equation?
\[
\text{A)} \{ 1,2 \} \quad
\text{B) } \{ -1, 2 \} \quad
\text{C) } \{ -2, -1 \} \quad
\text{D) } \{ -2 \} \quad
\text{E) } \{ -1 \}
\]
Solution:
\[ \frac { \displaystyle \frac{x}{x \ – \ 1}+ \displaystyle \frac{1}{1 \ – \ x} }{x} = \displaystyle \frac{x}{2 \ – \ x} \Rightarrow \frac{ \displaystyle \frac{x}{x \ – \ 1} \ – \ \displaystyle \frac{1}{x \ – \ 1} }{x} = \displaystyle \frac{x}{2 \ – \ x} \]
\[ \Rightarrow \frac{\displaystyle\frac{x – 1}{x – 1}}{x} = \frac{x}{2 – x} \]
\[ \Rightarrow \frac{1}{x} = \frac{x}{2 \ – \ x} \]
\[\Rightarrow x^2 = 2 \ – \ x \]
\[\Rightarrow x^2 +x \ – \ 2 = 0 \]
\[\Rightarrow (x \ – \ 1) \cdot (x+2 )= 0 \]
\( \Rightarrow x = 1 \) or \( \Rightarrow x = -2 \) is found. Here, since the denominator of the fraction \( \displaystyle \frac{x}{x-1 } \) becomes \(0 \) when \( x = 1 \), the number \( 1\) cannot be an element of the solution set. Therefore,
\[ S = \{ -2 \} \]
\(\textbf{Answer: D} \)
Question 3
\[ x = \frac{3}{m+1} \quad \text{and} \quad y = \frac{2m+3 }{m+1 } \]
What is $y$ expressed in terms of $x$?
\[
\text{A)} \frac{2+x}{5} \quad
\text{B) } \frac{6+x}{3} \quad
\text{C) } \frac{2x+1}{3} \quad
\text{D) } \frac{x+1}{5} \quad
\text{E) } x \ – \ 3
\]
Solution:
\[ y= \frac{2m+ 3 }{m+1 } = \frac{2 (m+1 ) +1}{m+1 } = 2+ \frac{1}{m+1} \]
\[ x = \frac{3}{m+1 } \Rightarrow \frac{x}{3} = \frac{1}{m+1 } \quad \text{and since this is true,} \]
\[y =2 + \frac{x}{3} \Rightarrow y = \frac{6+x }{3} \]
\(\textbf{Answer: B} \)
Question 4
\[ \frac{1+ \displaystyle\frac{x}{x \ – \ 1} }{\displaystyle\frac{1}{x \ – \ 1}+ \displaystyle\frac{1}{x+1} } = x+a \]
If the solution set of the equation is \( \displaystyle -\frac{1}{5} \), what is the value of the real number $a$?
\[
\text{A)} 1 \quad
\text{B) } 2 \quad
\text{C) } 3 \quad
\text{D) } 4 \quad
\text{E) } 5
\]
Solution:
\[ \frac{1+ \displaystyle\frac{x}{x \ – \ 1} }{\displaystyle\frac{1}{x \ – \ 1}+ \displaystyle\frac{1}{x+1} } = x+a \Rightarrow \frac{ \displaystyle\frac{2x \ – \ 1}{x \ – \ 1} }{\displaystyle\frac{2x}{(x \ – \ 1)(x+1)} } =x +a \]
\[ \Rightarrow \frac{(2x \ – \ 1) (x+1)}{2x} = x+a \]
\[\Rightarrow 2x^2+ x \ – \ 1 = 2x^2+ 2ax \]
\[ \Rightarrow x \ – \ 1 = 2ax \]
Since $S =$ \( \{ -\frac{1}{5} \} \), substituting the value \( x= -\frac{1}{5} \) into the equality above gives:
\[ – \frac{1}{5} \;- \;1 = 2a \;(-\frac{1}{5} ) \]
\[\Rightarrow a= 3 \]
\(\textbf{Answer: C} \)
Question 5
Given that \(a \neq b \):
\[ \frac{x}{a} + \frac{a \ – \ 1}{2} = \frac{x}{b} + \frac{b \ – \ 1}{2} \]
What value of $x$ satisfies the equation?
\[
\text{A) } a+b \quad
\text{B) } ab \quad
\text{C) } \frac{a+b}{2} \quad
\text{D) } \frac{-ab}{2} \quad
\text{E) } \frac{ab}{2}
\]
Solution:
\[\frac{x}{a} + \frac{a \ – \ 1}{2} = \frac{x}{b} + \frac{b \ – \ 1}{2} \]
\[ \Rightarrow \frac{x}{a} \;- \; \frac{x}{b} = \frac{b \ – \ 1}{2} \;-\; \frac{a \ – \ 1}{2} \]
\[ \Rightarrow \frac{x (b \ – \ a)}{ab} = \frac{b \ – \ a}{2} \Rightarrow x = \frac{ab}{2} \]
\(\textbf{Answer: E} \)
Question 6
\[ \displaystyle \frac{6}{4 \ – \ \Large \frac{3}{2 + \frac{1}{x+1}}} = 2 \]
What is the value of $x$?
\[
\text{A) } 0 \quad
\text{B) } 1 \quad
\text{C) } 2 \quad
\text{D) } 3 \quad
\text{E) } 4
\]
Solution:
\[ \frac{6}{4 \ – \ \frac{3}{2 + \frac{1}{x+1}}} = 2 \Rightarrow 4 \ – \ \frac{3}{2 + \frac{1}{x+1}} = 3 \]
\[ \frac{3}{2 + \frac{1}{x+1}} = 1 \Rightarrow 2 + \frac{1}{x+1} = 3 \]
\[ \frac{1}{x+1} = 1 \Rightarrow x+1 = 1 \Rightarrow x = 0 \]
\(\textbf{Answer: A} \)
Warning:
For \(a, b \in R \), in equalities of the form \(ax+ b = 0 \):
- If \( a= 0 \) and \( b= 0 \), then since \( 0 \cdot x + 0 = 0 \), every real number substituted for $x$ satisfies this equality.
- If \( a = 0 \) and \( b \neq 0\), then since \( 0 \cdot x + b=0 \), no real number substituted for $x$ can satisfy this equality.
Question 7
For \(a, b \in R \),
\[ \frac{2x \ – \ 1}{3} = \frac{x \ – \ a}{b} \]
given that this equality is satisfied by every real number, what is the sum \(a + b \)?
\[
\text{A) } 1 \quad
\text{B) } 2 \quad
\text{C) } 3 \quad
\text{D) } 4 \quad
\text{E) } 5
\]
Solution:
\[ \frac{2x \ – \ 1}{3} = \frac{x \ – \ a}{b} \Rightarrow 3 (x \ – \ a) = b (2x \ – \ 1) \]
\[ \Rightarrow (3 \ – \ 2b)x +b \ – \ 3a = 0\]
For this equality to be satisfied by every real number value of \( x \), we must have \( 3 \ – \ 2b =0 \) and \( b \ – \ 3a=0 \). From here,
\[ b = \frac{3}{2 }, \quad a = \frac{1}{2} \] and
\[a + b = \frac{3}{2} + \frac{1}{2} =2 \]
\(\textbf{Answer: B} \)
Question 8
For \(a, b \in R \),
\[ 2ax \ – \ b^2 = \frac{x \ – \ b^2+1}{3} \]
given that there is no real number value that satisfies this equality, what is the value of \(a \)?
\[
\text{A) } 2 \quad
\text{B) } 1 \quad
\text{C) } \frac{1}{2} \quad
\text{D) } \frac{1}{3} \quad
\text{E) } \frac{1}{6}
\]
Solution:
\[ 2ax \ – \ b^2 = \frac{x \ – \ b^2+1}{3} \Rightarrow 3 (2ax \ – \ b^2) = x \ – \ b^2+1 \]
\[ \Rightarrow (6a – 1)x -2b^2 -1 = 0 \]
For there to be no real number value of \( x \) that satisfies this equality,
\[6a – 1 = 0 \] and \[-2b^2-1 \neq 0 \] must hold. From here,
\[ a = \frac{1}{6} , \quad b^2 \neq -\frac{1}{2} \]
\(\textbf{Answer: E} \)
Question 9
For \(a, b \in R \),
\[ (1 \ – \ a)x + \frac{1}{x} = \frac{bx \ – \ c^2}{x} \]
given that there is no real number value of \(x\) that satisfies this equality, what is the sum \(a + b \)?
\[
\text{A) } 0 \quad
\text{B) } 1 \quad
\text{C) } 2 \quad
\text{D) } 3 \quad
\text{E) } 4
\]
Solution:
\[ (1 \ – \ a)x + \frac{1}{x} = \frac{bx-c^2}{x} \Rightarrow (1 \ – \ a) x^2 +1 = bx \ – \ c^2 \]
\[ \Rightarrow (1 \ – \ a)x^2 \ – \ bx + 1 +c^2 = 0 \]
For there to be no real number value that satisfies this equality,
\[1-a = 0 \] and \[ -b= 0 \] and \[ 1+ c^2 \neq 0 \] must hold.
From here,
\( a=1 \), \(- b=0\), and \( 1 +c^2 \neq 0\) must hold.
Thus,
\( a =1 \), \( b=0 \), and \(c^2 \neq -1\). Therefore, \( a+b= 1 + 0 = 1 \)
\(\textbf{Answer: B} \)
Question 10
\[ \frac{2x+1}{2} = \frac{3x \ – \ 2}{3} +2 \]
What is the solution set of the equation?
\[
\text{A) } \{ 1,3\} \quad
\text{B) } \{ 0,2\} \quad
\text{C) } \{ -1,2\} \quad
\text{D) } \{ 1\} \quad
\text{E) } \{ \emptyset \} \]
Solution:
\[ \frac{2x+1}{2} = \frac{3x \ – \ 2}{3} +2 \]
\[ \Rightarrow \frac{2x}{2}+ \frac{1}{2} = \frac{3x}{3} – \frac{2}{3} +2 \]
\[ \Rightarrow x \ – \ x + \frac{1}{2} + \frac{2}{3} \ – \ 2 = 0 \]
\[ 0 \cdot x \;- \; \frac{5}{6 } =0 \]
Therefore, \( S = \emptyset \)
\(\textbf{Answer: E} \)
Question 11
\[ 1+ \frac{x \ – \ 5}{2} = \frac{x \ – \ 3}{2} \]
What is the solution set of the equation?
\[
\text{A) } \{ 0\} \quad
\text{B) } \{ -1\} \quad
\text{C) } \{ 1\} \quad
\text{D) } R \quad
\text{E) } \{ \emptyset \}
\]
Solution:
\[1+ \frac{x \ – \ 5}{2} = \frac{x \ – \ 3}{2} \Rightarrow \frac{2 + x \ – \ 5}{2} = \frac{x \ – \ 3}{2} \]
\[\Rightarrow x \ – \ 3 = x \ – \ 3 \]
\[ x\ – \ x \ – \ 3+3 =0 \]
\[0 \cdot x + 0 =0 \]
Therefore, \( S = R \).
\(\textbf{Answer: D} \)
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