# Problems on Linear equation in one variable

This chapter is a collection of important problems related to linear equation in one variable.

We have already discussed different method of solving linear equation. Click the below red link to learn about the same.

## Problems on linear equation in one variable

Example 01
Solve the linear equation and find value of x.

\mathtt{\Longrightarrow \ 5\ +\ 20x\ =\ 4\ ( 2+3x)}

Solution
\mathtt{\Longrightarrow \ 5\ +\ 20x\ =\ 8\ +\ 12x}\\\ \\ \mathtt{\Longrightarrow \ 20x-12x\ =\ 8-5}\\\ \\ \mathtt{\Longrightarrow \ 8x\ =\ 3}\\\ \\ \mathtt{\Longrightarrow \ x=\frac{3}{8}}

Hence, x = 3/8 is the solution.

Example 02
Find the value of a.

\mathtt{\Longrightarrow \ 2( a+11) \ =\ 7\ ( a-3)}

Solution
\mathtt{\Longrightarrow \ 2a\ +22\ =\ 7a\ -21}\\\ \\ \mathtt{\Longrightarrow \ 2a-7a\ =\ -21-22}\\\ \\ \mathtt{\Longrightarrow \ -5a\ =\ -43}\\\ \\ \mathtt{\Longrightarrow \ a\ =\ \frac{-43}{-5}}\\\ \\ \mathtt{\Longrightarrow \ a\ =\frac{43}{5}}

Hence, value of a is 43/5.

Example 03
Solve linear equation and find value of x.

\mathtt{\Longrightarrow \ \frac{13}{5} +6x\ =\ \frac{2}{3}}

Solution
\mathtt{\Longrightarrow \ 6x\ =\ \frac{2}{3} -\frac{13}{5}}\\\ \\ \mathtt{\Longrightarrow \ 6x\ =\ \frac{10-39}{15}}\\\ \\ \mathtt{\Longrightarrow \ 6x\ =\ \frac{-29}{15}}

Cross multiplying number 6 towards the right we get;

\mathtt{\Longrightarrow \ x\ =\ \frac{-29}{15\times 6}}\\\ \\ \mathtt{\Longrightarrow \ x\ =\ \frac{-29}{90}}

Hence, value of x is -29/90.

Example 04
Find the value of x.

\mathtt{\Longrightarrow \frac{2x+5}{7} \ =\ \frac{8x-3}{5}}

Solution
Doing the cross multiplication.

\mathtt{\Longrightarrow \ 5\ ( 2x+5) \ =\ 7\ ( 8x-3)}\\\ \\ \mathtt{\Longrightarrow \ 10x\ +\ 25\ =\ 56x-21}\\\ \\ \mathtt{\Longrightarrow \ 25+21=56x-10x}\\\ \\ \mathtt{\Longrightarrow \ 46\ =\ 46x}\\\ \\ \mathtt{\Longrightarrow \ x\ =\ \frac{46}{46}}\\\ \\ \mathtt{\Longrightarrow \ x=1}

Hence, x = 1 is the solution of given expression.

Example 05
Solve linear equation and find value of y.

\mathtt{\Longrightarrow \frac{y+10}{y-9} \ =\ \frac{12}{13}}

Solution
Doing cross multiplication, we get;

\mathtt{\Longrightarrow \ 13\ ( y\ +\ 10) \ =\ 12\ ( y\ -\ 9)}\\\ \\ \mathtt{\Longrightarrow \ 13y+130\ =\ 12y-\ 108}\\\ \\ \mathtt{\Longrightarrow \ 13y-12y\ =-108-130}\\\ \\ \mathtt{\Longrightarrow \ y\ =\ -238}

Hence, y = -238 is the solution of given linear equation.

Example 06
Solve the below linear equation

\mathtt{\Longrightarrow 5( x+3) -2x\ =\ 16\ ( x+1)}

Solution
\mathtt{\Longrightarrow \ 5x+\ 15\ -\ 2x\ =\ 16x+16}\\\ \\ \mathtt{\Longrightarrow \ 3x+15=\ 16x+\ 16}\\\ \\ \mathtt{\Longrightarrow \ 3x-16x=\ 16-15}\\\ \\ \mathtt{\Longrightarrow \ -13x\ =\ 1}\\\ \\ \mathtt{\Longrightarrow \ x\ =\ \frac{-1}{13}}

Hence, x = -1/13 is the solution of given linear equation.

Example 07
Solve the linear equation.

\mathtt{\Longrightarrow \frac{x}{3} -\frac{5}{3} -\frac{x-3}{4} =\ \frac{3}{4} +1}

Solution
\mathtt{\Longrightarrow \frac{x}{3} -\frac{5}{3} -\frac{x-3}{4} =\ \frac{3}{4} +1}\\\ \\ \mathtt{\Longrightarrow \ \frac{x}{3} -\frac{x-3}{4} -\frac{5}{3} =\frac{3+4}{4}}\\\ \\ \mathtt{\Longrightarrow \ \frac{4x-\ 3( x-3)}{12} -\frac{5}{3} =\frac{7}{4}}\\\ \\ \mathtt{\Longrightarrow \ \frac{4x-3x+9}{12} =\ \frac{7}{4} +\frac{5}{3}}\\\ \\ \mathtt{\Longrightarrow \ \frac{x+9}{12} =\frac{21+20}{12}}

Cancelling number 12 on both sides.

\mathtt{\Longrightarrow \ \frac{x+9}{\cancel{12}} =\frac{21+20}{\cancel{12}}}\\\ \\ \mathtt{\Longrightarrow \ x+\ 9\ =\ 41}\\\ \\ \mathtt{\Longrightarrow \ x\ =\ 41-9}\\\ \\ \mathtt{\Longrightarrow \ x\ =\ 32}

Hence, the value of x is 32.

Example 08
Solve the linear equation and find value of x.

\mathtt{\Longrightarrow ( x\ +2)( x-3) =( x-5)( x-6)}

Solution
Multiplying the factors we get;

\mathtt{\Longrightarrow \ x^{2} -3x+2x-6=x^{2} -6x-5x+30}\\\ \\ \mathtt{\Longrightarrow \ \ \cancel{x^{2}} -3x+2x-6=\cancel{x^{2}} -6x-5x+30}\\\ \\ \mathtt{\Longrightarrow \ -x-6\ =-11x\ +\ 30}\\\ \\ \mathtt{\Longrightarrow \ -x+11x=30+6}\\\ \\ \mathtt{\Longrightarrow \ 10x\ =\ 36}\\\ \\ \mathtt{\Longrightarrow \ x=\frac{36}{10}}

The fraction can be further simplified by dividing numerator and denominator by 2.

\mathtt{\Longrightarrow \ x=\frac{36\div 2}{10\div 2}}\\\ \\ \mathtt{\Longrightarrow \ x\ =\ \frac{18}{5}}

Hence, 18/5 is the value of x.

Example 09
Solve the expression and find value of x.

\mathtt{\Longrightarrow ( x\ -2)( x-3) \ =\ ( x-7)^{2}}

Solution
Referring the formula;
\mathtt{( a-b)^{2} =a^{2} +b^{2} -2ab}

Applying the formula in main expression;

\mathtt{\Longrightarrow \ x^{2} -3x-2x+6\ =\ x^{2} +49-14x}\\\ \\ \mathtt{\Longrightarrow \ x^{2} -5x+6\ =\ x^{2} +49-14x}\\\ \\ \mathtt{\Longrightarrow \ \cancel{x^{2}} -5x+6\ =\cancel{x^{2}} \ +49-14x}\\\ \\ \mathtt{\Longrightarrow \ 14x-5x=\ 49-6}\\\ \\ \mathtt{\Longrightarrow \ 9x\ =\ 43}\\\ \\ \mathtt{\Longrightarrow \ x\ =\ \frac{43}{9}}

Hence, x = 43/9 is the solution.