DAV Class 7 Maths Chapter 7 Worksheet 1

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DAV Class 7 Maths Chapter 7 Worksheet 1

Linear Equations in One Variable Worksheet 1


1. Solve the following equations and check your answers.

(1) \( \quad 5x - 2 = 18\)

Solution:

\begin{align*} 5x - 2 &= 18 \\ 5x &= 18 + 2 \\ 5x &= 20 \\ x &= \frac{\cancel{20}^4}{\cancel5_1} \\ \\ \color{red} x & \, \color{red} = 4 \\ \\ &\color{Orange} \text{Check} \\ \color{magenta} \text{LHS} & = 5x - 2 \\ & = 5(4) - 2 \\ & = 20 - 2 \\ \color{magenta} \text{LHS} & = 18 \\ \\ \color{magenta} \text{RHS} & = 18 \\ \\ \text{LHS } &= \text{ RHS}\\ \end{align*}

Answer \(\color{red} {x = 4} \)

(2) \(\quad { \displaystyle\frac{1}{4}}y + {\displaystyle \frac{1}{2}} = 5\)

Solution:

\begin{align*} \frac{1}{4}y + \frac{1}{2} &= 5 \\ \\ \frac{1}{4}y &= 5 - \frac{1}{2} \\ \\ \frac{1}{4}y &= \frac{10 - 1}{2} \\ \\ \frac{1}{4}y &= \frac{9}{2} \\ \\ y &= \frac{9}{\cancel2_1} \times \cancel{4}^2 \\ \\ \color{red} y & \, \color{red} = 18 \\ \\ &\color{Orange} \text{Check} \\ \color{magenta} \text{LHS} & = \frac{1}{4}y + \frac{1}{2} \\ \\ & = \left(\frac{1}{\cancel4_2} \times \cancel{18}^9 \right) + \frac{1}{2} \\ \\ & =\frac{9}{2} + \frac{1}{2} \\ \\ & =\frac{9+1}{2} \\ \\ & =\frac{\cancel{10}^5}{\cancel2_1} \\ \\ \color{magenta} \text{LHS} & =5 \\ \\ \color{magenta} \text{RHS} & = 5 \\ \\ \text{LHS } &= \text{ RHS}\\\end{align*}

Answer \(\color{red} {y = 18} \)

(3) \(\quad { \displaystyle \frac{3}{5}}x - 6 = 3\)

Solution:

\begin{align*} \frac{3}{5}x - 6 &= 3 \\ \\ \frac{3}{5}x &= 3 + 6 \\ \\ \frac{3}{5}x &= 9 \\ x &= \cancel{9}^3 \times \frac{5}{\cancel3_1} \\ \color{red} x & \, \color{red} = 15 \\ \\ &\color{Orange} \text{Check} \\ \color{magenta} \text{LHS} & = \frac{3}{5}x - 6 \\ \\ & = \left(\frac{3}{\cancel5_1} \times \cancel{15}^3 \right) - 6 \\ \\ & = (3 \times 3) - 6 \\ & = 9 - 6 \\ \color{magenta} \text{LHS} & = 3 \\ \\ \color{magenta} \text{RHS} & = 3 \\ \\ \text{LHS } &= \text{ RHS}\\\end{align*}

Answer \(\color{red} {x = 15} \)

(4) \(\quad 3x + {\displaystyle \frac{1}{5}} = 2 - x \)

Solution:

\begin{align*} 3x + \frac{1}{5} &= 2 - x \\ \\ 3x + x &= 2 - \frac{1}{5} \\ \\ 4x &= \frac{10 - 1}{5} \\ \\ 4x &= \frac{9}{5} \\ \\ x &= \frac{9}{5} \times \frac{1}{4} \\ \\ \color{red} x & \, \color{red} = \frac{9}{20} \\ \\&\color{Orange} \text{Check} \\ \color{magenta} \text{LHS} & = 3x + {\frac{1}{5}} \\ \\ & = 3 \left( \frac{9}{20}\right) + \frac{1}{5} \\ \\ & = \frac{27}{20} + \frac{1 {\color{green}\times 4}}{5 {\color{green}\times 4}} \\ \\ & = \frac{27 + 4}{20} \\ \\ & = \frac{31}{20} \\ \\\color{magenta} \text{RHS} & = 2-x \\ \\ & = 2 - \frac{9}{20} \\ \\ & = \frac{2 {\color{green}\times 20}}{1 {\color{green}\times 20}} - \frac{9}{20} \\ \\ & =\frac{40 - 9}{20} \\ \\ & =\frac{31}{20} \\ \\ \text{LHS } &= \text{ RHS}\\ \end{align*}

Answer \(\color{red} {x = \displaystyle \frac{9}{20}} \)

(5) \(\quad 8x + 5 = 6x - 5\)

Solution:

\begin{align*} 8x + 5 &= 6x - 5 \\ 8x - 6x &= -5 - 5 \\ 2x &= -10 \\ \\ x &= \frac{-10}{2} \\ \\ \color{red} x & \, \color{red} = -5 \\ \\ &\color{Orange} \text{Check} \\ \end{align*} \begin{align*} \begin{array}{l|l} \color{magenta} \text{LHS} & \color{magenta} \text{RHS} \\ \hline = 8x + 5 & =6x - 5 \\ = 8 (-5) + 5 & = 6 (-5) - 5 \\ = -40 + 5 & =-30 - 5 \\ = -35 & =-35 \\ \end{array} \\ \end{align*}\begin{align*} \text{LHS} = \text{RHS} \end{align*}

Answer \(\color{red} x = -5 \)

(6) \(\quad 9z - 13 = 11z + 27 \)

Solution:

\begin{align*} 9z - 13 &= 11z + 27 \\ 9z - 11z &= 27 + 13 \\ -2z &= 40 \\ -z &= \frac{\cancel{40}^{20}}{\cancel2_1} \\ -z &= 20 \\ \color{red} z & \, \color{red} = -20 \\ \\ &\color{Orange} \text{Check} \\ \end{align*} \begin{align*} \begin{array}{l|l} \color{magenta} \text{LHS} & \color{magenta} \text{RHS} \\ \hline = 9z - 13 & =11z + 27 \\ = 9 (-20) - 13 & =11 (-20) + 27 \\ = -180 - 13 & =-220 + 27 \\ = -193 & =-193 \\ \end{array} \\ \end{align*} \begin{align*} \text{LHS} = \text{RHS} \end{align*}

Answer \(\color{red} z = - 20 \)

(7) \(\quad { \displaystyle \frac{7}{y}} +1 = 29 \)

Solution:

\begin{align*} \frac{7}{y} + 1 &= 29 \\ \\ \frac{7}{y} &= 29 - 1 \\ \\ \frac{7}{y} &= 28 \\ \\ 7 &= 28y \\ \frac{7}{28} &= y \\ \\ \frac{1}{4}&=y \\ \\ \color{red} y & \, \color{red} = \frac{1}{4} \\ \\&\color{Orange} \text{Check} \\ \color{magenta} \text{LHS} & = { \frac{7}{y}} + 1 \\ \\ & = { \frac{7}{\left({\frac{1}{4}}\right)}} + 1 \\ \\ & = \left(7 \times {{\frac{4}{1}}}\right) + 1 \\ \\ & = 28 + 1 \\ \color{magenta} \text{LHS} & = 29 \\ \\ \color{magenta} \text{RHS} & = 29 \\ \\\text{LHS} &= \text{RHS} \end{align*}

Answer \(\color{red} y = \displaystyle \frac{1}{4} \)

(8) \(\quad {\displaystyle \frac{3}{5}}x + {\displaystyle \frac{2}{5}} = 1 \)

Solution:

\begin{align*} \frac{3}{5}x + \frac{2}{5} &= 1 \\ \\ \frac{3}{5}x &= 1 - \frac{2}{5} \\ \\ \frac{3}{5}x &= \frac{1 {\color{green}\times 5}}{1 {\color{green}\times 5}} - \frac{2}{5} \\ \\ \frac{3}{5}x &= \frac{5-2}{5} \\ \\ \frac{3}{5}x &= \frac{3}{5} \\ \\ x &= \frac{3}{5} \times \frac{5}{3} \\ \\ \color{red} x & \, \color{red} = 1 \\ \\&\color{Orange} \text{Check} \\ \color{magenta} \text{LHS} & = {\frac{3}{5}x + \frac{2}{5}} \\ \\ & = \frac{3}{5} (1) + \frac{2}{5} \\ \\ & = \frac{3}{5} + \frac{2}{5} \\ \\ & = \frac{3+2}{5} \\ \\ & = \frac{5}{5} \\ \\ \color{magenta} \text{LHS} & = 1 \\ \\ \color{magenta} \text{RHS} & = 1 \\ \\\text{LHS} &= \text{RHS} \end{align*}

Answer \(\color{red} x = 1 \)

(9) \(\quad 4y -2 = {\displaystyle \frac{1}{5}}\)

Solution:

\begin{align*} 4y - 2 &= \frac{1}{5} \\ \\ 4y &= \frac{1}{5} + 2 \\ \\ 4y &= \frac{1}{5} + \frac{2 {\color{green}\times 5}}{1 {\color{green}\times 5}} \\ \\ 4y &= \frac{1 + 10}{5} \\ \\ 4y &= \frac{11}{5} \\ \\ y &= \frac{11}{5} \times \frac{1}{4} \\ \\\color{red} y & \, \color{red} = \frac{11}{20} \\ \\&\color{Orange} \text{Check} \\ \color{magenta} \text{LHS} & = 4y - 2 \\ \\ & = \cancel4^1 \left({\frac{11}{\cancel{20}_5}}\right) - 2 \\ \\ & = \frac{11}{5} - 2 \\ \\ & = \frac{11}{5} - \frac{2 {\color{green}\times 5}}{1 {\color{green}\times 5}} \\ \\ & = \frac{11 - 10}{5} \\ \\ \color{magenta} \text{LHS} & = \frac{1}{5} \\ \\\color{magenta} \text{RHS} & = \frac{1}{5} \\ \\ \text{LHS} &= \text{RHS} \end{align*}

Answer \(\color{red} y = \displaystyle \frac{11}{20} \)

(10) \(\quad {\displaystyle \frac{x}{2}} + {\displaystyle\frac{x}{4}} = 12\)

Solution:

\begin{align*} \frac{x}{2} + \frac{x}{4} &= 12 \\ \\ \frac{x {\color{green}\times 2} }{2 {\color{green}\times 2}} + \frac{x}{4} &= 12 \\ \\ \frac{2x + x}{4} &= 12 \\ \\ \frac{3x}{4} &= 12 \\ \\ x &= \cancel{12}^4 \times \frac{4}{\cancel3_1} \\ \\ \color{red} x & \, \color{red} = 16 \\ \\ &\color{Orange} \text{Check} \\ \color{magenta} \text{LHS} & = {\frac{x}{2} + \frac{x}{4}} \\ \\ &= {\frac{\cancel{16}^8}{\cancel2_1} + \frac{\cancel{16}^4}{\cancel4_1}} \\ \\ & = 8 + 4 \\ \color{magenta} \text{LHS} & = 12 \\ \\\color{magenta} \text{RHS} & = 12 \\ \\ \text{LHS} &= \text{RHS} \end{align*}

Answer \(\color{red} x = 16 \)

(11) \(\quad {\displaystyle \frac{2}{5}}z = {\displaystyle \frac{3}{8}}z + {\displaystyle \frac{7}{20}}\)

Solution:

\begin{align*} \frac{2}{5}z &= \frac{3}{8}z + \frac{7}{20} \\ \\ \frac{2}{5}z - \frac{3}{8}z &= \frac{7}{20} \\ \\ \frac{2 {\color{green}\times 8}}{5{\color{green}\times 8}}z - \frac{3{\color{green}\times 5}}{8 {\color{green}\times 5}}z &= \frac{7}{20} \\ \\ \frac{16z - 15z}{40} &= \frac{7}{20} \\ \\ \frac{1}{40}z &= \frac{7}{20} \\ \\ z &= \frac{7}{\cancel{20}_1} \times \cancel{40}^2 \\ \color{red} z & \, \color{red} = 14 \\ \\&\color{Orange} \text{Check} \\ \color{magenta} \text{LHS} & = {\frac{2}{5}}z \\ \\ & = {\frac{2}{5}} \times 14 \\ \\ & = {\frac{28}{5}} \\ \\ \color{magenta} \text{RHS} & = \frac{3}{8}z + \frac{7}{20} \\ \\ & = {\frac{3}{\cancel8_4}} (\cancel{14}^7) + {\frac{7}{20}} \\ \\ & = \frac{21}{4} + \frac{7}{20} \\ \\ & = \frac{21 {\color{green}\times 5}}{4 {\color{green}\times 5}} + \frac{7}{20} \\ \\ & = \frac{105+7}{20} \\ \\ & = \frac{\cancel{112}^{28}}{\cancel{20}_5} \\ \\ & = {\frac{28}{5}} \\ \\ \text{LHS} &= \text{RHS} \end{align*}

Answer \(\color{red} z = 14 \)

(12) \(\quad {\displaystyle \frac{2}{5}}y - {\displaystyle \frac{5}{8}}y = {\displaystyle \frac{5}{12}}\)

Solution:

\begin{align*} \frac{2y}{5} - \frac{5y}{8} &= \frac{5}{12} \\ \\ \frac{2y {\color{green}\times 8}}{5{\color{green}\times 8}} - \frac{5y {\color{green}\times 5}}{8 {\color{green}\times 5}} &= \frac{5}{12} \\ \\ \frac{16y - 25y}{40} &= \frac{5}{12} \\ \\ -\frac{9y}{40} &= \frac{5}{12} \\ \\ -y &= \frac{5}{\cancel{12}_3} \times \frac{\cancel{40}^{10}}{9} \\ \\ -y &= \frac{50}{27} \\ \\ \color{red} y & \, \color{red} = \frac{-50}{27} \\ \\ &\color{Orange} \text{Check} \\ \color{magenta} \text{LHS} & = {\frac{2}{5}}y - {\frac{5}{8}}y \\ \\ &= \frac{2}{\cancel5_1} \times \left(-\frac{\cancel{50}^{10}}{27}\right) - \frac{5}{\cancel8_4} \times \left(-\frac{\cancel{50}^{25}}{27}\right) \\ \\ & = -\frac{20}{27} + \frac{125}{108} \\ \\ & = -\frac{20 {\color{green}\times 4}}{27 {\color{green}\times 4}} + \frac{125}{108} \\ \\ & = \frac{-80 + 125}{108} \\ \\ & = \frac{\cancel{45}^5}{\cancel{108}_{12}} \\ \\ \color{magenta} \text{LHS} & = \frac{5}{12}\\ \\\color{magenta} \text{RHS} & = \frac{5}{12} \\ \\\text{LHS} &= \text{RHS} \end{align*}

Answer \(\color{red} y = \displaystyle \frac{-50}{27} \)

(13) \(\quad 3x + 2(x + 2) = 20 - (2x - 5)\)

Solution:

\begin{align*} 3x + 2(x + 2) &= 20 - (2x - 5) \\ 3x + 2x + 4 &= 20 - 2x + 5 \\ 5x + 4 &= 25 - 2x \\ 5x + 2x &= 25 - 4 \\ 7x &= 21 \\ x &= \frac{\cancel{21}^3}{\cancel7_1} \\ \color{red} x & \, \color{red} = 3 \\ \\ &\color{Orange} \text{Check} \\ \color{magenta} \text{LHS} & = 3x + 2(x + 2) \\ & = 3(3) + 2(3 + 2) \\ & = 9 + 2(5) \\ & = 9 + 10 \\ & = 19 \\ \\ \color{magenta} \text{RHS} & = 20 - (2x - 5) \\ & = 20 - [2(3) - 5] \\ & = 20 - (6 - 5) \\ & = 20 - (1)\\ & = 20 - 1\\ & = 19 \\ \\ \text{LHS} &= \text{RHS} \end{align*}

Answer \(\color{red} x = 3 \)

(14) \(\quad 13(y - 4) - 3(y -9) = 5(y + 4)\)

Solution:

\begin{align*} 13(y - 4) - 3(y - 9) &= 5(y + 4) \\ 13y - 52 - 3y + 27 &= 5y + 20 \\ 13y - 3y - 5y &= 20 + 52 - 27 \\ 5y &= 45 \\ y &= \frac{\cancel{45}^9}{\cancel5_1} \\ \color{red} y & \, \color{red} = 9 \\ \\ &\color{Orange} \text{Check} \\ \color{magenta} \text{LHS} & = 13(y - 4) - 3(y - 9) \\ & = 13(9 - 4) - 3(9 - 9) \\ & = 13(5) - 3(0) \\ & = 65 - 0 \\ & = 65 \\ \\ \color{magenta} \text{RHS} & = 5(y + 4) \\ & = 5(9 + 4) \\ & = 5(13) \\ & = 65 \\ \\ \text{LHS} &= \text{RHS} \end{align*}

Answer \(\color{red} y = 9 \)

(15) \(\quad (2z - 7) - 3(3z + 8) = 4z-9\)

Solution:

\begin{align*} (2z - 7) - 3(3z + 8) &= 4z - 9 \\ 2z - 7 - 9z - 24 &= 4z - 9 \\ -7z - 31 &= 4z - 9 \\ -7z - 4z &= -9 + 31 \\ -11z &= 22 \\ -z &= \frac{\cancel{22}^2}{\cancel{11}_1} \\ -z &= 2 \\ \color{red} z & \, \color{red} = -2 \\ \\ &\color{Orange} \text{Check} \\ \color{magenta} \text{LHS} & = (2z - 7) - 3(3z + 8) \\ & = [2(-2) - 7] - 3[3(-2) + 8] \\ & = (-4 - 7) - 3(-6 + 8) \\ & = -11 - 6 \\ & = -17 \\ \\ \color{magenta} \text{RHS} & = 4z - 9 \\ & = 4(-2) - 9 \\ & = -8 - 9 \\ & = -17 \\ \\ \text{LHS} &= \text{RHS} \end{align*}

Answer \(\color{red} z = -2 \)

(16) \(\quad 4(2y - 3) + 5(3y - 4) = 14\)

Solution:

\begin{align*} 4(2y - 3) + 5(3y - 4) &= 14 \\ 8y - 12 + 15y - 20 &= 14 \\ 23y - 32 &= 14 \\ 23y &= 14 + 32 \\ 23y &= 46 \\ y &= \frac{\cancel{46}^2}{\cancel{23}_1} \\ \color{red} y & \, \color{red} = 2 \\ \\ &\color{Orange} \text{Check} \\ \color{magenta} \text{LHS} & = 4(2y - 3) + 5(3y - 4) \\ & = 4[2(2) - 3] + 5[3(2) - 4] \\ & = 4(4 - 3) + 5(6 - 4) \\ & = 4(1) + 5(2) \\ & = 4 + 10 \\ \color{magenta} \text{LHS}& = 14 \\ \\ \color{magenta} \text{RHS}& = 14 \\ \\ \text{LHS} &= \text{RHS} \end{align*}

Answer \(\color{red} y = 2 \)

(17) \(\quad \displaystyle{\frac{x}{2} - \frac{x}{3} = \frac{x}{4} + \frac{1}{2}} \)

Solution:

\begin{align*} \frac{x}{2} - \frac{x}{3} &= \frac{x}{4} + \frac{1}{2} \\ \\ \frac{x}{2} - \frac{x}{3} - \frac{x}{4} &= \frac{1}{2} \\ \\ \frac{x {\color{green}\times 6}}{2 {\color{green}\times 6}} - \frac{x {\color{green}\times 4}}{3 {\color{green}\times 4}} - \frac{x {\color{green}\times 3}}{4 {\color{green}\times 3}} &= \frac{1}{2} \\ \\ \frac{6x - 4x -3x}{12} &= \frac{1}{2} \\ \\ \frac{-x}{12} &= \frac{1}{2} \\ \\ -x &= \frac{\cancel{12}^6}{\cancel2_1} \\ \\ -x &= 6 \\ \color{red} x & \, \color{red} = -6 \\ \\ &\color{Orange} \text{Check} \\ \color{magenta} \text{LHS} &= \frac{x}{2} - \frac{x}{3} \\ \\ &= \frac{-\cancel6^3}{\cancel2_1} - \frac{-\cancel6^2}{\cancel3_1} \\ \\ &= -3 -(-2) \\ &= -3 + 2 \\ &= -1 \\ \\ \color{magenta} \text{RHS} &= \frac{x}{4} + \frac{1}{2} \\ \\ &= \frac{-\cancel6^3}{\cancel4_2} + \frac{1}{2} \\ \\ &= \frac{-3}{2} + \frac{1}{2} \\ \\ &= \frac{-3+1}{2} \\ \\ &= \frac{-\cancel2^1}{\cancel2_1} \\ \\ &= -1 \\ \\ \text{LHS} &= \text{RHS} \end{align*}

Answer \(\color{red} x = -6 \)

(18) \(\quad z - \displaystyle{\frac{2z}{3} + \frac{z}{2}} = 5 \)

Solution:

\begin{align*} z - \frac{2z}{3} + \frac{z}{2} &= 5 \\ \\ \frac{z {\color{green}\times 6}}{1 {\color{green}\times 6}} - \frac{2z {\color{green}\times 2}}{3 {\color{green}\times 2}} + \frac{z {\color{green}\times 3}}{2 {\color{green}\times 3}} &= 5 \\ \\ \frac{6z - 4z + 3z}{6} &= 5 \\ \\ \frac{5z}{6} &= 5 \\ \\ z &= \cancel5^1 \times \frac{6}{\cancel5_1} \\ \color{red} z & \, \color{red} = 6 \\ \\ &\color{Orange} \text{Check} \\ \color{magenta} \text{LHS} &= z - \frac{2z}{3} + \frac{z}{2} \\ \\ &= 6 - \frac{2 \times \cancel6^2}{\cancel3_1} + \frac{\cancel6^3}{\cancel2_1} \\ \\ &= 6 - 4 + 3 \\ &= 2 + 3 \\ &= 5 \\ \\ \color{magenta} \text{RHS} &= 5 \\ \\ \text{LHS} &= \text{RHS} \end{align*}

Answer \(\color{red} z = 6 \)

(19) \(\quad \displaystyle{\frac{6y + 1}{2}} + 1 = {\frac{7y - 3}{3}} \)

Solution:

\begin{align*} \frac{6y + 1}{2} + 1 &= \frac{7y - 3}{3} \\ \\ \frac{6y + 1}{2} + \frac{1 {\color{green}\times 2}}{1 {\color{green}\times 2}} &= \frac{7y - 3}{3} \\ \\ \frac{6y + 1 + 2}{2} &= \frac{7y - 3}{3} \\ \\ \frac{6y + 3}{2} &= \frac{7y - 3}{3} \\ \\ 3(6y + 3) &= 2(7y - 3) \\ 18y + 9 &= 14y - 6 \\ 18y - 14y &= -6 - 9 \\ 4y &= -15 \\ \\ \color{red} y & \, \color{red} = \frac{-15}{4} \\ \\ &\color{Orange} \text{Check} \\\color{magenta} \text{LHS} &= \frac{6y + 1}{2} + 1 \\ \\ &= \frac{\cancel6^3\left(\frac{-15}{\cancel4_2}\right) + 1}{2} + 1 \\ \\ &= \frac{\frac{-45}{2} + 1}{2} + 1 \\ \\ &= \frac{\frac{-45}{2} + \frac{1 {\color{green}\times 2}}{1 {\color{green}\times 2}}}{2} + 1 \\ \\ &= \frac{\frac{-45+2}{2} }{2} + 1 \\ \\ &= \frac{\frac{-43}{2} }{2} + 1 \\ \\ &= \frac{-43 }{2 \times 2} + 1 \\ \\ &= \frac{-43 }{4} + 1 \\ \\ &= \frac{-43 }{4} + \frac{1 {\color{green}\times 4}}{1 {\color{green}\times 4}} \\ \\ &= \frac{-43 + 4}{4} \\ \\ &= \frac{-39}{4} \\ \\\color{magenta} \text{RHS} &= \frac{7y - 3}{3} \\ \\ &= \frac{7(\frac{-15}{4}) - 3}{3} \\ \\ &= \frac{\frac{-105}{4} - 3}{3} \\ \\ &= \frac{\frac{-105}{4} - \frac{3 {\color{green}\times 4}}{1 {\color{green}\times 4}}}{3} \\ \\ &= \frac{\frac{-105 -12}{4}}{3} \\ \\ &= \frac{\frac{-117}{4}}{3} \\ \\ &= \frac{-\cancel{117}^{39}}{4 \times \cancel3_1} \\ \\ &= \frac{-39}{4} \\ \\ \text{LHS} &= \text{RHS} \end{align*}

Answer \(\color{red} y = \displaystyle -\frac{15}{4} \)

(20) \(\quad \displaystyle{\frac{6x - 2}{5}} = {\frac{2x - 1}{3} - \frac{1}{3}} \)

Solution:

\begin{align*} \frac{6x - 2}{5} &= \frac{2x - 1}{3} - \frac{1}{3} \\ \\ \frac{6x - 2}{5} &= \frac{2x - 1 - 1}{3} \\ \\ \frac{6x - 2}{5} &= \frac{2x - 2}{3} \\ \\ 3(6x - 2) &= 5(2x - 2) \\ 18x - 6 &= 10x - 10 \\ 18x - 10x &= -10 + 6 \\ 8x &= -4 \\ x &= \frac{-\cancel4^1}{\cancel8_2} \\ \\ \color{red} x & \, \color{red} = \frac{-1}{2} \\ \\ &\color{Orange} \text{Check} \\ \color{magenta} \text{LHS} &= \frac{6x - 2}{5} \\ \\ &= \frac{\cancel6^3 \left(\frac{-1}{\cancel2_1}\right) - 2}{5} \\ \\ &= \frac{-3 - 2}{5} \\ \\ &= \frac{-\cancel5^1}{\cancel5_1} \\ \\ &= -1 \\ \\\color{magenta} \text{RHS} &= \frac{2x - 1}{3} - \frac{1}{3} \\ \\ &= \frac{\cancel2 \left(\frac{-1}{\cancel2} \right) - 1}{3} - \frac{1}{3} \\ \\ &= \frac{-1 - 1}{3} - \frac{1}{3} \\ \\ &= \frac{-2}{3} - \frac{1}{3} \\ \\ &= \frac{-2-1}{3} \\ \\ &= \frac{-3}{3} \\ \\ &= -1 \\ \\ \text{LHS} &= \text{RHS} \end{align*}

Answer \(\color{red} x =\displaystyle -\frac{1}{2} \)

(21) \(\quad \displaystyle {\frac{z - 1}{3}} = 1 + {\frac{z - 2}{4}} \)

Solution:

\begin{align*} \frac{z - 1}{3} &= 1 + \frac{z - 2}{4} \\ \\ \frac{z - 1}{3} - \frac{z - 2}{4} &= 1 \\ \\ \frac{(z - 1) {\color{green}\times 4}}{3 {\color{green}\times 4}} - \frac{(z - 2) {\color{green}\times 3}}{4 {\color{green}\times 3}} &= 1 \\ \\ \frac{4(z - 1) - 3(z - 2)}{12} &= 1 \\ \\ \frac{4z - 4 - 3z + 6}{12} &= 1 \\ \\ \frac{z + 2}{12} &= 1 \\ \\ z + 2 &= 12 \\ z &= 12-2 \\ \color{red} z & \, \color{red} = 10 \\ \\ &\color{Orange} \text{Check} \\ \color{magenta} \text{LHS} &= \frac{z - 1}{3} \\ \\ &= \frac{10 - 1}{3} \\ \\ &= \frac{\cancel9^3}{\cancel3_1} \\ \\ &= 3 \\ \\ \color{magenta} \text{RHS} &= 1 + \frac{z - 2}{4} \\ \\ &= 1 + \frac{10 - 2}{4} \\ \\ &= 1 + \frac{\cancel8^2}{\cancel4_1} \\ &= 1 + 2 \\ &= 3 \\ \\ \text{LHS} &= \text{RHS} \end{align*}

Answer \(\color{red} z = 10 \)

(22) \(\quad 2x - 3 = \displaystyle {\frac{3}{10}}(5x - 12) \)

Solution:

\begin{align*} 2x - 3 &= \frac{3}{10}(5x - 12) \\ \\ 10(2x - 3) &= 3(5x - 12) \\ 20x - 30 &= 15x - 36 \\ 20x - 15x &= -36 + 30 \\ 5x &= -6 \\ \\ \color{red} x & \, \color{red} = \frac{-6}{5} \\ \\ &\color{Orange} \text{Check} \\ \color{magenta} \text{LHS} &= 2x - 3 \\ \\ &= 2 \left(\frac{-6}{5} \right) - 3 \\ \\ &= \frac{-12}{5} - \frac{3 {\color{green}\times 5}}{1 {\color{green}\times 5}} \\ \\ &= \frac{-12-15}{5} \\ \\ &= \frac{-27}{5} \\ \\ \color{magenta} \text{RHS} &= \frac{3}{10}(5x - 12) \\ \\ &= \frac{3}{10} \times\left[\cancel5 \left(\frac{-6}{\cancel5} \right) - 12 \right] \\ \\ &= \frac{3}{10} \times (-6 - 12) \\ \\ &= \frac{3}{\cancel{10}_5}\times (-\cancel{18}^9) \\ \\ &= \frac{3 \times (-9)}{5} \\ \\ &= \frac{-27}{5} \\ \\ \text{LHS} &= \text{RHS} \end{align*}

Answer \(\color{red} x = \displaystyle -\frac{6}{5} \)

(23) \(\quad 3(y - 3) = 5(2y + 1)\)

Solution:

\begin{align*} 3(y - 3) &= 5(2y + 1) \\ 3y - 9 &= 10y + 5 \\ 3y - 10y &= 5 + 9 \\ -7y &= 14 \\ \\ -y & = \frac{\cancel{14}^2}{\cancel7_1} \\ \\ -y & = 2 \\ \color{red} y & \, \color{red} = -2 \\ \\ &\color{Orange} \text{Check} \\ \color{magenta} \text{LHS} &= 3(y - 3) \\ &= 3(-2 - 3) \\ &= 3(-5) \\ &= -15 \\ \\ \color{magenta} \text{RHS} &= 5(2y + 1) \\ \\ &= 5[2(-2) + 1] \\ \\ &= 5(-4 + 1) \\ \\ &= 5(-3) \\ \\ &= -15 \\ \\ \text{LHS} &= \text{RHS} \end{align*}

Answer \(\color{red} y = -2 \)

(24) \(\quad 0.6x + 0.8 = 0.28x + 1.16\)

Solution:

\begin{align*} 0.6x + 0.8 &= 0.28x + 1.16 \\ 0.6x - 0.28x &= 1.16 - 0.8 \\ 0.32x &= 0.36 \\ \\ x &= \frac{0.36 {\color{green}\times 100}}{0.32 {\color{green}\times 100}} \\ \\ x &= \frac{\cancel{36}^9}{\cancel{32}_8} \\ \\ x &= \frac{9}{8} \\ \\ \color{red} x & \, \color{red} = 1.125 \\ \\ &\color{Orange} \text{Check} \\ \color{magenta} \text{LHS} &= 0.6x + 0.8 \\ &= 0.6 \times (1.125) + 0.8 \\ &= 0.675 + 0.8 \\ &= 1.475 \\ \\ \color{magenta} \text{RHS} &= 0.28x + 1.16 \\ &= 0.28 \times (1.125) + 1.16 \\ &= 0.315 + 1.16 \\ &= 1.475 \\ \\ \text{LHS} &= \text{RHS} \end{align*}

Answer \(\color{red} x = 1.125 \)

(25) \(\quad 2.4(3 - x) - 0.6(2x - 3) = 0\)

Solution:

\begin{align*} 2.4(3 - x) - 0.6(2x - 3) &= 0 \\ 7.2 - 2.4x - 1.2x + 1.8 &= 0 \\ 7.2 + 1.8 - 2.4x - 1.2x &= 0 \\ 9 - 3.6x &= 0 \\ 3.6x &= 9 \\ x &= \frac{9 {\color{green}\times 10}}{3.6 {\color{green}\times 10}} \\ \\ x &= \frac{\cancel9^1 \times \cancel{10}^5}{{\cancel{36}_{\cancel4}}_2} \\ \\ x &= \frac{5}{2} \\ \\ \color{red} x & \, \color{red} = 2.5 \\ \\ \end{align*} \begin{align*} &\color{Orange} \text{Check} \\ \color{magenta} \text{LHS} &= 2.4(3 - x) - 0.6(2x - 3) \\ &= 2.4(3 - 2.5) - 0.6[2(2.5) - 3] \\ &= 2.4(0.5) - 0.6(5 - 3) \\ &= 1.2 - 0.6(2) \\ &= 1.2 - 1.2 \\ &= 0 \\ \\ \color{magenta} \text{RHS} &= 0 \\ \\ \text{LHS} &= \text{RHS} \end{align*}

Answer \(\color{red} x = 2.5 \)