question_answer 18)

Solve the following equations and check your results: 1.            \[3x=2x+18\]     2.            \[5t-3=3t-5\] 3.            \[5x+9=5+3x\] 4.            \[4z+3=6+2z\] 5.            \[2x-1=14-x\] 6.            \[8x+4=3(x-1)+7\] 7.            \[x=\frac{4}{5}\,(x+10)\] 8.            \[\frac{2x}{3}+1=\frac{7x}{15}+3\] 9.            \[2y+\frac{5}{3}=\frac{26}{3}-y\] 10.          \[3m=5m-\frac{8}{5}\]

Answer:

                1. \[3x=2x+18\]                 We have \[3x=2x+18\]  | Transposing 2x to LHS \[3x-2x=18\] \[\Rightarrow \]               \[x=18\] This is the required solution. Check:   LHS \[=3x=3\times 18=54\] RHS \[=2x+18=2\times 18+18\] \[=36+18=54\] Therefore, LHS = RHS     | as desired 2.            \[5t-3=3t-5\]                 We have \[5t-3=3t-5\]                   |transposing 3t to LHS and ? 3 to RHS \[5t-3t=-5+3\] \[\Rightarrow \]               \[2t=-2\] \[\Rightarrow \]               \[t=-\frac{2}{2}=-1\]                                       | Dividing both sides by 2 This is the required solution.           Check: LHS \[=5(-1)-3\] \[=-5-3=-8\] RHS \[=3t-5=3(-1)-5~~~~\] \[=-35=-8~~~~~~~~\] Therefore, LHS = RHS                     | as desired 3.            \[5x+9=5+3x\] We have \[5x+9=5+3x\]                | Transposing \[3x\] to LHS and 9 to RHS                 \[5x-3x=5-9\] \[\Rightarrow \]               \[2x=-4\] \[\Rightarrow \]               \[x=-\frac{4}{2}=-2\]                                      | Dividing both sides by 2 This is the required solution. Check:   LHS \[=5(-2)+9\] \[=-10+9=-1\] RHS \[=5+3(-2)\] \[=56=-1\] Therefore, LHS = RHS                     | as desired 4.            \[4z+3=6+2z\] We have \[4z+3=6+2z\] | Transposing \[2x\] to LHS and 3 to RHS \[4z-2z=6-3\] \[\Rightarrow \]               \[2z=3\] \[\Rightarrow \]               \[z=\frac{3}{2}\]                                                               | Dividing both sides by 2 This is the required solution. Check:  LHS \[=4\left( \frac{3}{2} \right)\,+3=6+3=9\]                 RHS \[=6+2z=6+2\left( \frac{3}{2} \right)\]                 \[=6+3=9\] Therefore, LHS = RHS                     | as desired 5.            \[2x-1=14-x\] We have \[2x-1=14-x\] \[\Rightarrow \]               \[2x+x=1+1\]                     | Transposing \[-x\] to LHS and \[-1\]to RHS \[\Rightarrow \]               \[3x=15\] \[\Rightarrow \]               \[x=\frac{15}{3}=5\]                       | Dividing both sides by 3 This is the required solution. Check:   LHS \[=2x-1\] \[=2(5)-1=10-1\] \[=9\] RHS \[=14-x\] \[=14-5=9\] Therefore, LHS = RHS     | as desired 6.            \[8x+4=3(x-1)+7\] We have \[8x+43=3(x-1)+7\] \[\Rightarrow \]               \[8x+4=3x-3+7\] \[\Rightarrow \]               \[8x+4=3x+4\] \[\Rightarrow \]               \[8x-3x=4-4\]     | Transposing \[3x\] to LHS and 4 to RHS \[\Rightarrow \]               \[5x=0\] \[\Rightarrow \]               \[x=\frac{0}{5}\,=0\]                      | Dividing both sides by 5 This is the required solution. Check: LHS \[=8x+4\] \[=8(0)+4=4\]                 RHS \[=3(x-1)+7\] \[=3(0-1)+7\] \[=3(-1)+7\] \[=-3+7=4\] Therefore, LHS = RHS     | as desired 7.            \[x=\frac{4}{5}\,(x+10)\]                 We have \[x=\frac{4}{5}\,(x+10)\] \[\Rightarrow \]               \[5x=4(x+10)\]                  |Multiplying both sides by 5 \[\Rightarrow \]               \[5x=4x+40\] \[\Rightarrow \]               \[5x-4x=40\]                      |Transposing \[4x\] to LHS \[\Rightarrow \]               \[x=40\]                 This is the required solution. Cheek: LHS = 40 RHS \[=\frac{4}{5}(x+10)\] \[=\frac{4}{5}(40+10)\] \[=\frac{4}{5}(50)\] \[=4(10)=40\] Therefore, LHS = RHS     |as desired 8.            \[\frac{2x}{3}+1=\frac{7x}{15}+3\]                 We have \[\frac{2x}{3}+1=\frac{7x}{15}+3\] \[\Rightarrow \]               \[\frac{2x}{3}-\frac{7x}{15}=3-1\]             |Transposing \[\frac{7x}{15}\] to LHS        and 1 to RHS \[\Rightarrow \]               \[\frac{2x}{3}-\frac{7x}{15}=2\] \[\Rightarrow \]               \[15\left( \frac{2x}{3}-\frac{7x}{15} \right)\,=2\times 15\]            |Multiplying both sides by 15 \[\Rightarrow \]               \[10x-7x=30\] \[\Rightarrow \]               \[3x=30\]             \[\Rightarrow \]               \[x=\frac{30}{3}=10\]                     |Dividing both sides by 3 This is the required solution. Check: LHS \[=\frac{2x}{3}\,+1=\frac{2}{3}(10)+1\] \[=\frac{20}{3}+1\] \[=\frac{20+3}{3}=\frac{23}{3}\] RHS \[=\frac{7x}{15}+3\] \[=\frac{7}{15}(10)+3=\frac{70}{15}+3\] \[=\frac{70\div 5}{15\div 5}\,+3=\frac{14}{3}+3\] \[=\frac{14+9}{3}\,=\frac{23}{3}\] Therefore, LHS = RHS 9.            \[2y+\frac{5}{3}\,=\frac{26}{3}-y\]                 We have \[2y+\frac{5}{3}=\frac{26}{3}-y\]           |Transposing \[-y\] to LHS and \[\frac{5}{3}\] to RHS \[2y+y=\frac{26}{3}-\frac{5}{3}\] \[\Rightarrow \]               \[3y=\frac{26-5}{3}\] \[\Rightarrow \]               \[3y=\frac{21}{3}\] \[\Rightarrow \]               \[3y=7\] \[\Rightarrow \]               \[y=\frac{7}{3}\]                                                               |Dividing both sides by 3 This is the required solution. Check: LHS \[=2\left( \frac{7}{3} \right)+\frac{5}{3}=\frac{14}{3}+\frac{5}{3}\] \[=\frac{14+5}{3}\,=\frac{19}{3}\] \[=\frac{26}{3}-\frac{7}{3}=\frac{26-7}{3}=\frac{19}{3}\] Therefore, \[LHS=RHS\]                | as desired 10.          \[3m=5m-\frac{8}{5}\]                 We have \[3m=5m-\frac{8}{5}\] \[\Rightarrow \]               \[3m-5m=-\frac{8}{5}\]                 |Transposing 5m to LHS \[\Rightarrow \]               \[-2m=-\frac{8}{5}\] \[\Rightarrow \]               \[m=\frac{-8}{5\times (-2)}=\frac{4}{5}\]              |Dividing both sides by ? 2 This is the required solution. Check: LHS \[=3\times \frac{4}{5}\,=\frac{12}{5}\] RHS\[=5\left( \frac{4}{5} \right)\,-\frac{8}{5}\,=4-\frac{8}{5}\] \[=\frac{20-8}{5}=\frac{12}{5}\] Therefore, LHS = RHS                     | as desired