Solve for $$ x $$ and $$ y $$ by using the method of substitution: $$ \begin{array}{lll} ext { (1) } x-y=8 & ext { and } 2 x+y=10 & ext { (2) } 3 x-y=2 ext { and } 7 x-2 y \ ext { (3) } 3 x+5 y=8 ext { and } x-2 y=-1 & ext { (4) } 7 x-3 y=41 ext { and } 3 x-y \ ext { Solve for } x ext { and } y ext { using either the elimination or substitution method: } \ \begin{array}{lll} ext { (1) } x+y=1 ext { and } x-2 y=1 & ext { (2) } & 3 x+2 y=2 ext { and } 5 x-2 ext { ) } \ ext { (3) } x+4 y=14 ext { and } 3 x+2 y=12 & ext { (4) } 2 y-3 x=7 ext { and } 4 y-5 x \ ext { (5) } 3 x+2 y=6 ext { and } 5 x+3 y=11 & \ ext { (6) } \frac{y-2}{3}-\frac{x-1}{2}=1 ext { and } \frac{y-4}{5}-\frac{x-3}{4}=1\end{array}\end{array} l $$

Question

Solve for $$ x $$ and $$ y $$ by using the method of substitution: $$ \begin{array}{lll}	ext { (1) } x-y=8 & 	ext { and } 2 x+y=10 & 	ext { (2) } 3 x-y=2 	ext { and } 7 x-2 y \ 	ext { (3) } 3 x+5 y=8 	ext { and } x-2 y=-1 & 	ext { (4) } 7 x-3 y=41 	ext { and } 3 x-y \ 	ext { Solve for } x 	ext { and } y 	ext { using either the elimination or substitution method: } \ \begin{array}{lll}	ext { (1) } x+y=1 	ext { and } x-2 y=1 & 	ext { (2) } & 3 x+2 y=2 	ext { and } 5 x-2 	ext { ) } \ 	ext { (3) } x+4 y=14 	ext { and } 3 x+2 y=12 & 	ext { (4) } 2 y-3 x=7 	ext { and } 4 y-5 x \ 	ext { (5) } 3 x+2 y=6 	ext { and } 5 x+3 y=11 & \ 	ext { (6) } \frac{y-2}{3}-\frac{x-1}{2}=1 	ext { and } \frac{y-4}{5}-\frac{x-3}{4}=1\end{array}\end{array} l $$

UpStudy AI · Accepted Answer

Solve the system of equations by following steps:
- step0: Solve using the substitution method:
  $$\left\{ \begin{array}{l}x+y=1\x-2y=1\end{array}ight.$$
- step1: Solve the equation:
  $$\left\{ \begin{array}{l}x=1-y\x-2y=1\end{array}ight.$$
- step2: Substitute the value of $$x:$$
  $$1-y-2y=1$$
- step3: Subtract the terms:
  $$1-3y=1$$
- step4: Move the constant to the right side:
  $$-3y=1-1$$
- step5: Subtract the terms:
  $$-3y=0$$
- step6: Change the signs:
  $$3y=0$$
- step7: Rewrite the expression:
  $$y=0$$
- step8: Substitute the value of $$y:$$
  $$x=1-0$$
- step9: Substitute back:
  $$x=1$$
- step10: Calculate:
  $$\left\{ \begin{array}{l}x=1\y=0\end{array}ight.$$
- step11: Check the solution:
  $$\left\{ \begin{array}{l}x=1\y=0\end{array}ight.$$
- step12: Rewrite:
  $$\left(x,yight) = \left(1,0ight)$$
Solve the system of equations $$ x-y=8; 2 x+y=10 $$.
Solve the system of equations by following steps:
- step0: Solve using the substitution method:
  $$\left\{ \begin{array}{l}x-y=8\2x+y=10\end{array}ight.$$
- step1: Solve the equation:
  $$\left\{ \begin{array}{l}x=8+y\2x+y=10\end{array}ight.$$
- step2: Substitute the value of $$x:$$
  $$2\left(8+yight)+y=10$$
- step3: Simplify:
  $$16+3y=10$$
- step4: Move the constant to the right side:
  $$3y=10-16$$
- step5: Subtract the numbers:
  $$3y=-6$$
- step6: Divide both sides:
  $$\frac{3y}{3}=\frac{-6}{3}$$
- step7: Divide the numbers:
  $$y=-2$$
- step8: Substitute the value of $$y:$$
  $$x=8-2$$
- step9: Calculate:
  $$x=6$$
- step10: Calculate:
  $$\left\{ \begin{array}{l}x=6\y=-2\end{array}ight.$$
- step11: Check the solution:
  $$\left\{ \begin{array}{l}x=6\y=-2\end{array}ight.$$
- step12: Rewrite:
  $$\left(x,yight) = \left(6,-2ight)$$
Solve the system of equations $$ 3 x+5 y=8; x-2 y=-1 $$.
Solve the system of equations by following steps:
- step0: Solve using the substitution method:
  $$\left\{ \begin{array}{l}3x+5y=8\x-2y=-1\end{array}ight.$$
- step1: Solve the equation:
  $$\left\{ \begin{array}{l}3x+5y=8\x=-1+2y\end{array}ight.$$
- step2: Substitute the value of $$x:$$
  $$3\left(-1+2yight)+5y=8$$
- step3: Simplify:
  $$-3+11y=8$$
- step4: Move the constant to the right side:
  $$11y=8+3$$
- step5: Add the numbers:
  $$11y=11$$
- step6: Divide both sides:
  $$\frac{11y}{11}=\frac{11}{11}$$
- step7: Divide the numbers:
  $$y=1$$
- step8: Substitute the value of $$y:$$
  $$x=-1+2	imes 1$$
- step9: Substitute back:
  $$x=-1+2$$
- step10: Calculate:
  $$x=1$$
- step11: Calculate:
  $$\left\{ \begin{array}{l}x=1\y=1\end{array}ight.$$
- step12: Check the solution:
  $$\left\{ \begin{array}{l}x=1\y=1\end{array}ight.$$
- step13: Rewrite:
  $$\left(x,yight) = \left(1,1ight)$$
Solve the system of equations $$ 3 x-y=2; 7 x-2 y=0 $$.
Solve the system of equations by following steps:
- step0: Solve using the substitution method:
  $$\left\{ \begin{array}{l}3x-y=2\7x-2y=0\end{array}ight.$$
- step1: Solve the equation:
  $$\left\{ \begin{array}{l}y=-2+3x\7x-2y=0\end{array}ight.$$
- step2: Substitute the value of $$y:$$
  $$7x-2\left(-2+3xight)=0$$
- step3: Simplify:
  $$x+4=0$$
- step4: Move the constant to the right side:
  $$x=0-4$$
- step5: Remove 0:
  $$x=-4$$
- step6: Substitute the value of $$x:$$
  $$y=-2+3\left(-4ight)$$
- step7: Calculate:
  $$y=-14$$
- step8: Calculate:
  $$\left\{ \begin{array}{l}x=-4\y=-14\end{array}ight.$$
- step9: Check the solution:
  $$\left\{ \begin{array}{l}x=-4\y=-14\end{array}ight.$$
- step10: Rewrite:
  $$\left(x,yight) = \left(-4,-14ight)$$
Solve the system of equations $$ 7 x-3 y=41; 3 x-y=0 $$.
Solve the system of equations by following steps:
- step0: Solve using the substitution method:
  $$\left\{ \begin{array}{l}7x-3y=41\3x-y=0\end{array}ight.$$
- step1: Solve the equation:
  $$\left\{ \begin{array}{l}7x-3y=41\y=3x\end{array}ight.$$
- step2: Substitute the value of $$y:$$
  $$7x-3	imes 3x=41$$
- step3: Simplify:
  $$-2x=41$$
- step4: Change the signs:
  $$2x=-41$$
- step5: Divide both sides:
  $$\frac{2x}{2}=\frac{-41}{2}$$
- step6: Divide the numbers:
  $$x=-\frac{41}{2}$$
- step7: Substitute the value of $$x:$$
  $$y=3\left(-\frac{41}{2}ight)$$
- step8: Calculate:
  $$y=-\frac{123}{2}$$
- step9: Calculate:
  $$\left\{ \begin{array}{l}x=-\frac{41}{2}\y=-\frac{123}{2}\end{array}ight.$$
- step10: Check the solution:
  $$\left\{ \begin{array}{l}x=-\frac{41}{2}\y=-\frac{123}{2}\end{array}ight.$$
- step11: Rewrite:
  $$\left(x,yight) = \left(-\frac{41}{2},-\frac{123}{2}ight)$$
Solve the system of equations $$ x+4 y=14; 3 x+2 y=12 $$.
Solve the system of equations by following steps:
- step0: Solve using the substitution method:
  $$\left\{ \begin{array}{l}x+4y=14\3x+2y=12\end{array}ight.$$
- step1: Solve the equation:
  $$\left\{ \begin{array}{l}x=14-4y\3x+2y=12\end{array}ight.$$
- step2: Substitute the value of $$x:$$
  $$3\left(14-4yight)+2y=12$$
- step3: Simplify:
  $$42-10y=12$$
- step4: Move the constant to the right side:
  $$-10y=12-42$$
- step5: Subtract the numbers:
  $$-10y=-30$$
- step6: Change the signs:
  $$10y=30$$
- step7: Divide both sides:
  $$\frac{10y}{10}=\frac{30}{10}$$
- step8: Divide the numbers:
  $$y=3$$
- step9: Substitute the value of $$y:$$
  $$x=14-4	imes 3$$
- step10: Calculate:
  $$x=2$$
- step11: Calculate:
  $$\left\{ \begin{array}{l}x=2\y=3\end{array}ight.$$
- step12: Check the solution:
  $$\left\{ \begin{array}{l}x=2\y=3\end{array}ight.$$
- step13: Rewrite:
  $$\left(x,yight) = \left(2,3ight)$$
Solve the system of equations $$ (y-2)/3-(x-1)/2=1; (y-4)/5-(x-3)/4=1 $$.
Solve the system of equations by following steps:
- step0: Solve using the substitution method:
  $$\left\{ \begin{array}{l}\frac{\left(y-2ight)}{3}-\frac{\left(x-1ight)}{2}=1\\frac{\left(y-4ight)}{5}-\frac{\left(x-3ight)}{4}=1\end{array}ight.$$
- step1: Remove the parentheses:
  $$\left\{ \begin{array}{l}\frac{y-2}{3}-\frac{x-1}{2}=1\\frac{y-4}{5}-\frac{x-3}{4}=1\end{array}ight.$$
- step2: Solve the equation:
  $$\left\{ \begin{array}{l}x=\frac{-7+2y}{3}\\frac{y-4}{5}-\frac{x-3}{4}=1\end{array}ight.$$
- step3: Substitute the value of $$x:$$
  $$\frac{y-4}{5}-\frac{\frac{-7+2y}{3}-3}{4}=1$$
- step4: Simplify:
  $$\frac{y-4}{5}+\frac{8-y}{6}=1$$
- step5: Multiply both sides of the equation by LCD:
  $$\left(\frac{y-4}{5}+\frac{8-y}{6}ight)	imes 30=1	imes 30$$
- step6: Simplify the equation:
  $$y+16=30$$
- step7: Move the constant to the right side:
  $$y=30-16$$
- step8: Subtract the numbers:
  $$y=14$$
- step9: Substitute the value of $$y:$$
  $$x=\frac{-7+2	imes 14}{3}$$
- step10: Calculate:
  $$x=7$$
- step11: Calculate:
  $$\left\{ \begin{array}{l}x=7\y=14\end{array}ight.$$
- step12: Check the solution:
  $$\left\{ \begin{array}{l}x=7\y=14\end{array}ight.$$
- step13: Rewrite:
  $$\left(x,yight) = \left(7,14ight)$$
Solve the system of equations $$ 3 x+2 y=2; 5 x-2 y=0 $$.
Solve the system of equations by following steps:
- step0: Solve using the substitution method:
  $$\left\{ \begin{array}{l}3x+2y=2\5x-2y=0\end{array}ight.$$
- step1: Solve the equation:
  $$\left\{ \begin{array}{l}x=\frac{2-2y}{3}\5x-2y=0\end{array}ight.$$
- step2: Substitute the value of $$x:$$
  $$5	imes \frac{2-2y}{3}-2y=0$$
- step3: Simplify:
  $$\frac{5\left(2-2yight)}{3}-2y=0$$
- step4: Multiply both sides of the equation by LCD:
  $$\left(\frac{5\left(2-2yight)}{3}-2yight)	imes 3=0	imes 3$$
- step5: Simplify the equation:
  $$10-16y=0$$
- step6: Move the constant to the right side:
  $$-16y=0-10$$
- step7: Remove 0:
  $$-16y=-10$$
- step8: Change the signs:
  $$16y=10$$
- step9: Divide both sides:
  $$\frac{16y}{16}=\frac{10}{16}$$
- step10: Divide the numbers:
  $$y=\frac{5}{8}$$
- step11: Substitute the value of $$y:$$
  $$x=\frac{2-2	imes \frac{5}{8}}{3}$$
- step12: Calculate:
  $$x=\frac{1}{4}$$
- step13: Calculate:
  $$\left\{ \begin{array}{l}x=\frac{1}{4}\y=\frac{5}{8}\end{array}ight.$$
- step14: Check the solution:
  $$\left\{ \begin{array}{l}x=\frac{1}{4}\y=\frac{5}{8}\end{array}ight.$$
- step15: Rewrite:
  $$\left(x,yight) = \left(\frac{1}{4},\frac{5}{8}ight)$$
Solve the system of equations $$ 2 y-3 x=7; 4 y-5 x=0 $$.
Solve the system of equations by following steps:
- step0: Solve using the substitution method:
  $$\left\{ \begin{array}{l}2y-3x=7\4y-5x=0\end{array}ight.$$
- step1: Solve the equation:
  $$\left\{ \begin{array}{l}x=\frac{-7+2y}{3}\4y-5x=0\end{array}ight.$$
- step2: Substitute the value of $$x:$$
  $$4y-5	imes \frac{-7+2y}{3}=0$$
- step3: Simplify:
  $$4y-\frac{5\left(-7+2yight)}{3}=0$$
- step4: Multiply both sides of the equation by LCD:
  $$\left(4y-\frac{5\left(-7+2yight)}{3}ight)	imes 3=0	imes 3$$
- step5: Simplify the equation:
  $$2y+35=0$$
- step6: Move the constant to the right side:
  $$2y=0-35$$
- step7: Remove 0:
  $$2y=-35$$
- step8: Divide both sides:
  $$\frac{2y}{2}=\frac{-35}{2}$$
- step9: Divide the numbers:
  $$y=-\frac{35}{2}$$
- step10: Substitute the value of $$y:$$
  $$x=\frac{-7+2\left(-\frac{35}{2}ight)}{3}$$
- step11: Calculate:
  $$x=-14$$
- step12: Calculate:
  $$\left\{ \begin{array}{l}x=-14\y=-\frac{35}{2}\end{array}ight.$$
- step13: Check the solution:
  $$\left\{ \begin{array}{l}x=-14\y=-\frac{35}{2}\end{array}ight.$$
- step14: Rewrite:
  $$\left(x,yight) = \left(-14,-\frac{35}{2}ight)$$
Solve the system of equations $$ 3 x+2 y=6; 5 x+3 y=11 $$.
Solve the system of equations by following steps:
- step0: Solve using the substitution method:
  $$\left\{ \begin{array}{l}3x+2y=6\5x+3y=11\end{array}ight.$$
- step1: Solve the equation:
  $$\left\{ \begin{array}{l}x=\frac{6-2y}{3}\5x+3y=11\end{array}ight.$$
- step2: Substitute the value of $$x:$$
  $$5	imes \frac{6-2y}{3}+3y=11$$
- step3: Simplify:
  $$\frac{5\left(6-2yight)}{3}+3y=11$$
- step4: Multiply both sides of the equation by LCD:
  $$\left(\frac{5\left(6-2yight)}{3}+3yight)	imes 3=11	imes 3$$
- step5: Simplify the equation:
  $$30-y=33$$
- step6: Move the constant to the right side:
  $$-y=33-30$$
- step7: Subtract the numbers:
  $$-y=3$$
- step8: Change the signs:
  $$y=-3$$
- step9: Substitute the value of $$y:$$
  $$x=\frac{6-2\left(-3ight)}{3}$$
- step10: Calculate:
  $$x=4$$
- step11: Calculate:
  $$\left\{ \begin{array}{l}x=4\y=-3\end{array}ight.$$
- step12: Check the solution:
  $$\left\{ \begin{array}{l}x=4\y=-3\end{array}ight.$$
- step13: Rewrite:
  $$\left(x,yight) = \left(4,-3ight)$$
Here are the solutions for each system of equations using the method of substitution:

1. For the equations $$ x - y = 8 $$ and $$ 2x + y = 10 $$:
   - The solution is $$ (x, y) = (6, -2) $$.

2. For the equations $$ 3x - y = 2 $$ and $$ 7x - 2y = 0 $$:
   - The solution is $$ (x, y) = (-4, -14) $$.

3. For the equations $$ 3x + 5y = 8 $$ and $$ x - 2y = -1 $$:
   - The solution is $$ (x, y) = (1, 1) $$.

4. For the equations $$ 7x - 3y = 41 $$ and $$ 3x - y = 0 $$:
   - The solution is $$ (x, y) = \left(-\frac{41}{2}, -\frac{123}{2}ight) $$ or approximately $$ (x, y) = (-20.5, -61.5) $$.

5. For the equations $$ x + y = 1 $$ and $$ x - 2y = 1 $$:
   - The solution is $$ (x, y) = (1, 0) $$.

6. For the equations $$ 3x + 2y = 2 $$ and $$ 5x - 2y = 0 $$:
   - The solution is $$ (x, y) = \left(\frac{1}{4}, \frac{5}{8}ight) $$ or approximately $$ (x, y) = (0.25, 0.625) $$.

7. For the equations $$ x + 4y = 14 $$ and $$ 3x + 2y = 12 $$:
   - The solution is $$ (x, y) = (2, 3) $$.

8. For the equations $$ 2y - 3x = 7 $$ and $$ 4y - 5x = 0 $$:
   - The solution is $$ (x, y) = \left(-14, -\frac{35}{2}ight) $$ or approximately $$ (x, y) = (-14, -17.5) $$.

9. For the equations $$ 3x + 2y = 6 $$ and $$ 5x + 3y = 11 $$:
   - The solution is $$ (x, y) = (4, -3) $$.

10. For the equations $$ \frac{y-2}{3} - \frac{x-1}{2} = 1 $$ and $$ \frac{y-4}{5} - \frac{x-3}{4} = 1 $$:
    - The solution is $$ (x, y) = (7, 14) $$.

These results provide the values of $$ x $$ and $$ y $$ for each system of equations.
Here are the solutions for each system of equations using substitution: 1. $$ (x, y) = (6, -2) $$ 2. $$ (x, y) = (-4, -14) $$ 3. $$ (x, y) = (1, 1) $$ 4. $$ (x, y) = \left(-\frac{41}{2}, -\frac{123}{2}ight) $$ or approximately $$ (-20.5, -61.5) $$ 5. $$ (x, y) = (1, 0) $$ 6. $$ (x, y) = \left(\frac{1}{4}, \frac{5}{8}ight) $$ or approximately $$ (0.25, 0.625) $$ 7. $$ (x, y) = (2, 3) $$ 8. $$ (x, y) = \left(-14, -\frac{35}{2}ight) $$ or approximately $$ (-14, -17.5) $$ 9. $$ (x, y) = (4, -3) $$ 10. $$ (x, y) = (7, 14) $$

Solve for and by using the method of substitution:

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