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{20}{5}\\ x &= 4 \\ \text{Verification} \\ \end{align*} \begin{align*} \begin{array}{l|l} \text{LHS} & \text{RHS} \\ \hline = 5x - 2 & =18 \\ = 5(4) - 2 & \\ = 20 - 2 & \\ = 18 & \\ \end{array} \\ \text{LHS } = \text{ RHS}\\ \end{align*}
\({\boxed{x = 4}}\) is a solution of the given equation(2) \(\quad {\frac{1}{4}}y + {\frac{1}{2}} = 5\)
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} \times \cancelto{2}4 \\ y &= 18 \\ \text{Verification} \\ \end{align*} \begin{align*} { \begin{array}{l|l} \text{LHS} & \text{RHS} \\ \hline \\ = \frac{1}{4}y + \frac{1}{2} & =5 \\ = \left(\frac{1}{4} \times 18\right) + \frac{1}{2} & \\ = \frac{18}{4} + \frac{1}{2} & \\ = \frac{9}{2} + \frac{1}{2} & \\ = \frac{10}{2} & \\ = 5 & \\ \end{array} \\ }\\ \text{LHS} = \text{RHS}\\ \end{align*}
\({\boxed{y = 18}}\) is a solution of the given equation(3) \(\quad {\frac{3}{5}}x - 6 = 3\)
Solution:
\begin{align*} \frac{3}{5}x - 6 &= 3 \\ \frac{3}{5}x &= 3 + 6 \\ \frac{3}{5}x &= 9 \\ x &= \cancelto{3}9 \times \frac{5}{\cancel3} \\ x &= 15 \\ \text{Verification} \end{align*} \begin{align*} { \begin{array}{l|l} \text{LHS} & \text{RHS} \\ \hline = \frac{3}{5}x - 6 & =3 \\ = \left(\frac{3}{5} \times 15\right) - 6 & \\ = \frac{45}{5} - 6 & \\ = 9 - 6 & \\ = 3 & \\ \end{array} \\ }\\ \text{LHS } = \text{ RHS} \end{align*}
\({\boxed{x = 15}}\) is a solution of the given equation(4) \(\quad 3x + {\frac{1}{5}} = 2 - x\)
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} \\ x &= \frac{9}{20} \\ \text{Verification} \\ \end{align*} \begin{align*} { \begin{array}{l|l} \text{LHS} & \text{RHS} \\ \hline = 3x + \frac{1}{5} & =2 - x \\ = \left(3 \times \frac{9}{20}\right) + \frac{1}{5} & =2 - \left(\frac{9}{20}\right) \\ = \frac{27}{20} + \frac{1}{5} & = 2 - \frac{9}{20} \\ = \frac{27 + 4}{20} & =\frac{40 - 9}{20} \\ = \frac{31}{20} & =\frac{31}{20} \\ \end{array} \\ } \\ \end{align*} \begin{align*} \text{LHS } = \text{ RHS} \end{align*}
\(\boxed{x = \frac{9}{20}}\) is a solution of the given equation(5) \(\quad 8x + 5 = 6x - 5\)
Solution:
\begin{align*} 8x + 5 &= 6x - 5 \\ 8x - 6x &= -5 - 5 \\ 2x &= -10 \\ x &= \frac{-10}{2}\\ x &= -5 \\ \text{Verification} \\ \end{align*} \begin{align*} \begin{array}{l|l} \text{LHS} & \text{RHS} \\ \hline = 8x + 5 & =6x - 5 \\ = 8 \times (-5) + 5 & =6 \times (-5) - 5 \\ = -40 + 5 & =-30 - 5 \\ = -35 & =-35 \\ \end{array} \\ \end{align*} \begin{align*} \text{LHS} = \text{RHS} \end{align*}
\(\boxed{x = -5}\) is a solution of the given equation(6) \(\quad 9z - 13 = 11z + 27 \)
Solution:
\begin{align*} 9z - 13 &= 11z + 27 \\ 9z - 11z &= 27 + 13 \\ -2z &= 40 \\ z &= \frac{40}{-2} \\ z &= -20 \\ \text{Verification} \\ \end{align*} \begin{align*} \begin{array}{l|l} \text{LHS} & \text{RHS} \\ \hline = 9z - 13 & =11z + 27 \\ = 9 \times (-20) - 13 & =11 \times (-20) + 27 \\ = -180 - 13 & =-220 + 27 \\ = -193 & =-193 \\ \end{array} \\ \end{align*} \begin{align*} \text{LHS} = \text{RHS} \end{align*}
\(\boxed{z = -20}\) is a solution of the given equation(7) \(\quad {\frac{7}{y}} +1 = 29 \)
Solution:
\begin{align*} \frac{7}{y} + 1 &= 29 \\ \frac{7}{y} &= 29 - 1 \\ \frac{7}{y} &= 28 \\ 28 y &= 7 \\ y &= \frac{7}{28} \\ y &= \frac{1}{4} \\ \text{Verification} \\ \end{align*} \begin{align*} \begin{array}{l|l} \text{LHS} & \text{RHS} \\ \hline \\ = { \frac{7}{y}} + 1 & =29 \\ = { \frac{7}{\left({\frac{1}{4}}\right)}} + 1 & \\ = \left(7 \times {{\frac{4}{1}}}\right) + 1 & \\ = 28 + 1 & \\ = 29 & \\ \end{array} \\ \end{align*} \begin{align*} \text{LHS} = \text{RHS} \end{align*}
\(\boxed{y = \frac{1}{4}}\) is a solution of the given equation(8) \(\quad {\frac{3}{5}}x + {\frac{2}{5}} = 1 \)
Solution:
\begin{align*} \frac{3}{5}x + \frac{2}{5} &= 1 \\ \frac{3}{5}x &= 1 - \frac{2}{5} \\ \frac{3}{5}x &= \frac{5}{5} - \frac{2}{5} \\ \frac{3}{5}x &= \frac{3}{5} \\ x &= \frac{3}{5} \times \frac{5}{3} \\ x &= 1 \\ \\ \text{Verification} \\ \end{align*} \begin{align*} \begin{array}{l|l} \text{LHS} & \text{RHS} \\ \hline = {\frac{3}{5}x + \frac{2}{5}} & =1 \\ = \frac{3}{5} \times (1) + \frac{2}{5} & \\ = { \frac{3}{5} + \frac{2}{5}} & \\ = { \frac{5}{5}} & \\ = 1 & \\ \end{array} \\ \end{align*} \begin{align*} \text{LHS} = \text{RHS} \end{align*}
\(\boxed{x = 1}\) is a solution of the given equation(9) \(\quad 4y -2 = {\frac{1}{5}}\)
Solution:
\begin{align*} 4y - 2 &= \frac{1}{5} \\ 4y &= \frac{1}{5} + 2 \\ 4y &= \frac{1}{5} + \frac{10}{5} \\ 4y &= \frac{11}{5} \\ y &= \frac{11}{5} \times \frac{1}{4} \\ y &= \frac{11}{20} \\ \\ \end{align*} \begin{align*} \text{Verification} \\ \begin{array}{l|l} \text{LHS} & \text{RHS} \\ \hline = 4y - 2 & ={\frac{1}{5}} \\ = 4 \times \left({\frac{11}{20}}\right) - 2 & \\ = \frac{44}{20} - 2 & \\ = {\frac{44 - 40}{20}} & \\ = {\frac{4}{20}} & \\ = {\frac{1}{5}} & \\ \end{array} \\ \end{align*} \begin{align*} \text{LHS} = \text{RHS} \end{align*}
\(\boxed{y = \frac{11}{20}}\) is a solution of the given equation(10) \(\quad {\frac{x}{2}} + {\frac{x}{4}} = 12\)
Solution:
\begin{align*} \frac{x}{2} + \frac{x}{4} &= 12 \\ \frac{2x + x}{4} &= 12 \\ \frac{3x}{4} &= 12 \\ 3x &= 12 \times 4 \\ 3x &= 48 \\ x &= \frac{48}{3} \\ x &= 16 \\ \\ \text{Verification} \\ \end{align*} \begin{align*} \begin{array}{l|l} \text{LHS} & \text{RHS} \\ \hline = {\frac{x}{2} + \frac{x}{4}} & =12 \\ = {\frac{16}{2} + \frac{16}{4}} & \\ = 8 + 4 & \\ = 12 & \\ \end{array} \\ \end{align*} \begin{align*} \text{LHS} = \text{RHS} \end{align*}
\(\boxed{x = 16}\) is a solution of the given equation(11) \(\quad {\frac{2}{5}}z = {\frac{3}{8}}z + {\frac{7}{20}}\)
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{16z - 15z}{40} &= \frac{7}{20} \\ \frac{1}{40}z &= \frac{7}{20} \\ z &= \frac{7}{\cancel{20}} \times \cancelto{2}{40} \\ z &= 14 \\ \\ \text{Verification} \\ \end{align*} \begin{align*} \begin{array}{l|l} \text{LHS} & \text{RHS} \\ \hline = {\frac{2}{5}}z & =\frac{3}{8}z + \frac{7}{20} \\ = {\frac{2}{5}}\times 14 & = ({\frac{3}{8}} \times 14) + {\frac{7}{20}} \\ = {\frac{28}{5}} & ={\frac{42}{8} + \frac{7}{20}} \\ & ={\frac{42}{8} + \frac{7}{20}} \\ & ={\frac{210 + 14}{40}} \\ & ={\frac{28}{5}} \\ \end{array} \\ \end{align*} \begin{align*} \text{LHS} = \text{RHS} \end{align*}
\(\boxed{z = 14}\) is a solution of the given equation(12) \(\quad {\frac{2}{5}}y - {\frac{5}{8}}y = {\frac{5}{12}}\)
Solution:
\begin{align*} \frac{2}{5}y - \frac{5}{8}y &= \frac{5}{12} \\ \left( \frac{2}{5} - \frac{5}{8} \right)y &= \frac{5}{12} \\ \left( \frac{16 - 25}{40} \right)y &= \frac{5}{12} \\ \left( -\frac{9}{40} \right)y &= \frac{5}{12} \\ y &= \frac{5}{12} \times \left( -\frac{40}{9} \right) \\ y &= -\frac{5 \times \cancelto{10}{40}}{\cancelto{3}{12} \times 9} \\ y &= -\frac{5 \times 10}{3 \times 9} \\ y &= -\frac{50}{27} \\ \\ \end{align*} \begin{align*} \text{Verification} \\ \end{align*} \begin{align*} \begin{array}{l|l} \text{LHS} & \text{RHS} \\ \hline = \frac{2}{5}y - \frac{5}{8}y & = \frac{5}{12} \\ = \frac{2}{\cancel5} \times \left(-\frac{\cancelto{10}{50}}{27}\right) - \frac{5}{\cancelto{4}8} \times \left(-\frac{\cancelto{25}{50}}{27}\right) & \\ = -\frac{20}{27} + \frac{125}{108} & \\ = \frac{-80 + 125}{108} & \\ = \frac{45}{108} & \\ = \frac{5}{12} & \\ \end{array} \\ \end{align*} \begin{align*} \text{LHS} = \text{RHS} \end{align*}
\(\boxed{y = -\frac{50}{27}}\) is a solution of the given equation(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{21}{7} \\ x &= 3 \\ \\ \text{Verification} \\ \end{align*} \begin{align*} \begin{array}{l|l} \text{LHS} & \text{RHS} \\ \hline = 3x + 2(x + 2) & = 20 - (2x - 5) \\ = 3(3) + 2(3 + 2) & = 20 - (2(3) - 5) \\ = 9 + 2(5) & = 20 - (6 - 5) \\ = 9 + 10 & = 20 - 1 \\ = 19 & = 19 \\ \end{array} \\ \end{align*} \begin{align*} \text{LHS} &= \text{RHS} \end{align*}
\(\boxed{x = 3}\) is a solution of the given equation(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{45}{5} \\ y &= 9 \\ \\ \text{Verification} \\ \end{align*} \begin{align*} \begin{array}{l|l} \text{LHS} & \text{RHS} \\ \hline = 13(y - 4) - 3(y - 9) & = 5(y + 4) \\ = 13(9 - 4) - 3(9 - 9) & = 5(9 + 4) \\ = 13(5) - 3(0) & = 5(13) \\ = 65 - 0 & = 65 \\ = 65 & \\ \end{array} \\ \end{align*} \begin{align*} \text{LHS} &= \text{RHS} \end{align*}
\(\boxed{y = 9}\) is a solution of the given equation(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{22}{-11} \\ z &= -2 \\ \\ \text{Verification} \\ \end{align*} \begin{align*} \begin{array}{l|l} \text{LHS} & \text{RHS} \\ \hline = (2z - 7) - 3(3z + 8) & = 4z - 9 \\ = [2(-2) - 7] - 3[3(-2) + 8] & = 4(-2) - 9 \\ = (-4 - 7) - 3(-6 + 8) & = -8 - 9 \\ = -11 - 3(2) & = -17 \\ = -11 - 6 & \\ = -17 & \\ \end{array} \\ \end{align*} \begin{align*} \text{LHS} &= \text{RHS} \end{align*}
\(\boxed{z = -2}\) is a solution of the given equation(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{46}{23} \\ y &= 2 \\ \\ \text{Verification} \\ \end{align*} \begin{align*} \begin{array}{l|l} \text{LHS} & \text{RHS} \\ \hline = 4(2y - 3) + 5(3y - 4) & = 14 \\ = 4(2(2) - 3) + 5(3(2) - 4) & \\ = 4(4 - 3) + 5(6 - 4) & \\ = 4(1) + 5(2) & \\ = 4 + 10 & \\ = 14 & \\ \end{array} \\ \end{align*} \begin{align*} \text{LHS} &= \text{RHS} \end{align*}
\(\boxed{y = 2}\) is a solution of the given equation(17) \(\quad {\frac{x}{2} - \frac{x}{3} = \frac{x}{4} + \frac{1}{2}} \)
Solution:
\begin{align*} \frac{x}{2} - \frac{x}{3} &= \frac{x}{4} + \frac{1}{2} \\ \frac{3x - 2x}{6} &= \frac{x + 2}{4} \\ \frac{x}{6} &= \frac{x + 2}{4} \\ 4x &= 6(x + 2) \\ 4x &= 6x + 12 \\ -2x &= 12 \\ x &= -6 \\ \end{align*} \begin{align*} \text{Verification} \\ \end{align*} \begin{align*} \begin{array}{l|l} \text{LHS} & \text{RHS} \\ \hline = \frac{x}{2} - \frac{x}{3} & = \frac{x}{4} + \frac{1}{2} \\ = \frac{-6}{2} - \frac{-6}{3} & = \frac{-6}{4} + \frac{1}{2} \\ = -3 + 2 & = -\frac{3}{2} + \frac{1}{2} \\ = -1 & = -1 \\ \end{array} \\ \end{align*} \begin{align*} \text{LHS} &= \text{RHS} \end{align*}
\(\boxed{x = -6}\) is a solution of the given equation(18) \(\quad z - {\frac{2z}{3} + \frac{z}{2}} = 5 \)
Solution:
\begin{align*} z - \frac{2z}{3} + \frac{z}{2} &= 5 \\ \frac{6z - 4z + 3z}{6} &= 5 \\ \frac{5z}{6} &= 5 \\ 5z &= 30 \\ z &= \frac{30}{5} \\ z &= 6 \\ \end{align*} \begin{align*} \text{Verification} \\ \end{align*} \begin{align*} \begin{array}{l|l} \text{LHS} & \text{RHS} \\ \hline = z - \frac{2z}{3} + \frac{z}{2} & = 5 \\ = 6 - \frac{2 \times 6}{3} + \frac{6}{2} & \\ = 6 - 4 + 3 & \\ = 5 & \\ \end{array} \\ \end{align*} \begin{align*} \text{LHS} &= \text{RHS} \end{align*}
\(\boxed{z = 6}\) is a solution of the given equation(19) \(\quad {\frac{6y + 1}{2}} + 1 = {\frac{7y - 3}{3}} \)
Solution:
\begin{align*} \frac{6y + 1}{2} + 1 &= \frac{7y - 3}{3} \\ \frac{6y + 1}{2} + \frac{2}{2} &= \frac{7y - 3}{3} \\ \frac{6y + 3}{2} &= \frac{7y - 3}{3} \\ \frac{6y + 3}{2} \times 6 &= \frac{7y - 3}{3} \times 6 \\ 18y + 9 &= 14y - 6 \\ 18y - 14y &= -6 - 9 \\ 4y &= -15 \\ y &= -\frac{15}{4} \\ \end{align*} \begin{align*} \text{Verification} \\ \end{align*} \begin{align*} \begin{array}{l|l} \text{LHS} & \text{RHS} \\ \hline = \frac{6y + 1}{2} + 1 & = \frac{7y - 3}{3} \\ = \frac{6(-\frac{15}{4}) + 1}{2} + 1 & = \frac{7(-\frac{15}{4}) - 3}{3} \\ = \frac{\frac{-45 + 2}{2}}{2} + 1 & = \frac{-\frac{105}{4} - 3}{3} \\ = \frac{-43}{2 \times 2} + 1 & = \frac{-\frac{105 - 12}{4}}{3} \\ = \frac{-43 + 4}{4} & = \frac{-117}{3 times 4} \\ = \frac{-39}{4} & = \frac{-39}{4} \\ \end{array} \\ \end{align*} \begin{align*} \text{LHS} &= \text{RHS} \end{align*}
\(\boxed{y = -\frac{15}{4}}\) is a solution of the given equation(20) \(\quad {\frac{6x - 2}{5}} = {\frac{2x - 1}{3} - \frac{1}{3}} \)
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{-4}{8} \\ x &= -\frac{1}{2} \\ \end{align*} \begin{align*} \text{Verification} \\ \end{align*} \begin{align*} \begin{array}{l|l} \text{LHS} & \text{RHS} \\ \hline = \frac{6x - 2}{5} & = \frac{2x - 1}{3} - \frac{1}{3} \\ = \frac{6(-\frac{1}{2}) - 2}{5} & = \frac{2(-\frac{1}{2}) - 1}{3} - \frac{1}{3} \\ = \frac{-3 - 2}{5} & = \frac{-1 - 1}{3} - \frac{1}{3} \\ = \frac{-5}{5} & = \frac{-2}{3} - \frac{1}{3} \\ = -1 & = \frac{-2 - 1}{3} \\ & = \frac{-3}{3} \\ & = -1 \\ \end{array} \\ \end{align*} \begin{align*} \text{LHS} &= \text{RHS} \end{align*}
\(\boxed{x = -\frac{1}{2}}\) is a solution of the given equation(21) \(\quad {\frac{z - 1}{3}} = 1 + {\frac{z - 2}{4}} \)
Solution:
\begin{align*} \frac{z - 1}{3} &= 1 + \frac{z - 2}{4} \\ \frac{z - 1}{3} - \frac{z - 2}{4} &= 1 \\ \frac{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 \\ z &= 10 \\ \end{align*} \begin{align*} \text{Verification} \\ \end{align*} \begin{array}{l|l} \text{LHS} & \text{RHS} \\ \hline = \frac{z - 1}{3} & = 1 + \frac{z - 2}{4} \\ = \frac{10 - 1}{3} & = 1 + \frac{10 - 2}{4} \\ = \frac{9}{3} & = 1 + \frac{8}{4} \\ = 3 & = 1 + 2 \\ & = 3 \\ \end{array} \begin{align*} \text{LHS} &= \text{RHS} \end{align*}
\(\boxed{z = 10}\) is a solution of the given equation(22) \(\quad 2x - 3 = {\frac{3}{10}}(5x - 12) \)
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 \\ x &= \frac{-6}{5} \\ \end{align*} \begin{align*} \text{Verification} \\ \end{align*} \begin{align*} \begin{array}{l|l} \text{LHS} & \text{RHS} \\ \hline = 2x - 3 & = \frac{3}{10}(5x - 12) \\ = 2\left(\frac{-6}{5}\right) - 3 & = \frac{3}{10}\left(5\left(\frac{-6}{5}\right) - 12\right) \\ = \frac{-12}{5} - 3 & = \frac{3}{10}\left(-6 - 12\right) \\ = \frac{-12 - 15}{5} & = \frac{3}{\cancelto{5}{10}} \times \frac{\cancelto{9}{-18}}{1} \\ = \frac{-27}{5} & = \frac{3 \times (-9)}{5 \times 1} \\ & = \frac{-27}{5} \\ \end{array} \\ \end{align*} \begin{align*} \text{LHS} &= \text{RHS} \end{align*}
\(\boxed{x = \frac{-6}{5}}\) is a solution of the given equation(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{14}{-7} \\ y &= -2 \\ \end{align*} \begin{align*} \text{Verification} \\ \end{align*} \begin{align*} \begin{array}{l|l} \text{LHS} & \text{RHS} \\ \hline = 3(y - 3) & = 5(2y + 1) \\ = 3(-2 - 3) & = 5(2(-2) + 1) \\ = 3(-5) & = 5(-4 + 1) \\ = -15 & = 5(-3) \\ & = -15 \\ \end{array} \\ \end{align*} \begin{align*} \text{LHS} = \text{RHS} \end{align*}
\(\boxed{y = -2}\) is a solution of the given equation(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}{0.32} \\ x &= \frac{0.36 \times 100}{0.32 \times 100} \\ x &= \frac{36}{32} \\ x &= \frac{9}{8} \\ x &= 1.125 \\ \end{align*} \begin{align*} \text{Verification} \\ \end{align*} \begin{align*} \begin{array}{l|l} \text{LHS} & \text{RHS} \\ \hline = 0.6x + 0.8 & = 0.28x + 1.16 \\ = (0.6 \times 1.125) + 0.8 & = (0.28\times 1.125) + 1.16 \\ = 0.675 + 0.8 & = 0.315 + 1.16 \\ = 1.475 & = 1.475 \\ \end{array} \\ \end{align*} \begin{align*} \text{LHS} = \text{RHS} \end{align*}
\(\boxed{x = 1.125}\) is a solution of the given equation(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}{3.6} \\ x &= \frac{9 \times 10}{3.6 \times 10} \\ x &= \frac{\cancelto5{90}}{\cancelto{2}{36}} \\ x &= \frac{5}{2} \\ x &= 2.5 \\ \end{align*} \begin{align*} \text{Verification} \\ \end{align*} \begin{align*} \begin{array}{l|l} \text{LHS} & \text{RHS} \\ \hline = 2.4(3 - x) - 0.6(2x - 3) & = 0 \\ = 2.4(3 - 2.5) - 0.6(2 \times 2.5 - 3) & \\ = 2.4(0.5) - 0.6(2) & \\ = 1.2 - 1.2 & \\ = 0 & = 0 \\ \end{array} \\ \end{align*} \begin{align*} \text{LHS} = \text{RHS} \end{align*}
\(\boxed{x = 2.5}\) is a solution of the given equation