answers: 1.. \( x+y=10 \) \[ x-y=2 \] 2. \( x+y=5 \) 3. \( 2 x+3 y=11 \) \( 4 y-2 x=10 \) 4. \[ 3 x+2 y=13 \] \[ 2 x+3 y=18 \] \[ 8 x=7-3 y \] 6. 5. \( 8 x=7-3 y \) \( 3 x+4 y=3 \) \[ 9 y=-20 x+13 \] \( 6 x=7-10 y \) 7. \( 2 y=3 x+3 \) 8. \( 3 y-4 x=4 \) \( 2 y=x \) \( 8 x-1=-2 y \) 9. \( 4 y=-x \) 10. \[ \begin{array}{l} 3 y+6=6 x \\ x=2 y-2 \end{array} \]

Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}3y+6=6x\\x=2y-2\end{array}\right.\) - step1: Substitute the value of \(x:\) \(3y+6=6\left(2y-2\right)\) - step2: Expand the expression: \(3y+6=12y-12\) - step3: Move the expression to the left side: \(3y-12y=-12-6\) - step4: Add and subtract: \(-9y=-12-6\) - step5: Add and subtract: \(-9y=-18\) - step6: Change the signs: \(9y=18\) - step7: Divide both sides: \(\frac{9y}{9}=\frac{18}{9}\) - step8: Divide the numbers: \(y=2\) - step9: Substitute the value of \(y:\) \(x=2\times 2-2\) - step10: Calculate: \(x=2\) - step11: Calculate: \(\left\{ \begin{array}{l}x=2\\y=2\end{array}\right.\) - step12: Check the solution: \(\left\{ \begin{array}{l}x=2\\y=2\end{array}\right.\) - step13: Rewrite: \(\left(x,y\right) = \left(2,2\right)\) Solve the system of equations \( x+y=10;x-y=2 \). Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}x+y=10\\x-y=2\end{array}\right.\) - step1: Solve the equation: \(\left\{ \begin{array}{l}x=10-y\\x-y=2\end{array}\right.\) - step2: Substitute the value of \(x:\) \(10-y-y=2\) - step3: Subtract the terms: \(10-2y=2\) - step4: Move the constant to the right side: \(-2y=2-10\) - step5: Subtract the numbers: \(-2y=-8\) - step6: Change the signs: \(2y=8\) - step7: Divide both sides: \(\frac{2y}{2}=\frac{8}{2}\) - step8: Divide the numbers: \(y=4\) - step9: Substitute the value of \(y:\) \(x=10-4\) - step10: Calculate: \(x=6\) - step11: Calculate: \(\left\{ \begin{array}{l}x=6\\y=4\end{array}\right.\) - step12: Check the solution: \(\left\{ \begin{array}{l}x=6\\y=4\end{array}\right.\) - step13: Rewrite: \(\left(x,y\right) = \left(6,4\right)\) Solve the system of equations \( 2 x+3 y=11;4 y-2 x=10 \). Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}2x+3y=11\\4y-2x=10\end{array}\right.\) - step1: Solve the equation: \(\left\{ \begin{array}{l}x=\frac{11-3y}{2}\\4y-2x=10\end{array}\right.\) - step2: Substitute the value of \(x:\) \(4y-2\times \frac{11-3y}{2}=10\) - step3: Simplify: \(7y-11=10\) - step4: Move the constant to the right side: \(7y=10+11\) - step5: Add the numbers: \(7y=21\) - step6: Divide both sides: \(\frac{7y}{7}=\frac{21}{7}\) - step7: Divide the numbers: \(y=3\) - step8: Substitute the value of \(y:\) \(x=\frac{11-3\times 3}{2}\) - step9: Calculate: \(x=1\) - step10: Calculate: \(\left\{ \begin{array}{l}x=1\\y=3\end{array}\right.\) - step11: Check the solution: \(\left\{ \begin{array}{l}x=1\\y=3\end{array}\right.\) - step12: Rewrite: \(\left(x,y\right) = \left(1,3\right)\) Solve the system of equations \( 3 x+2 y=13;2 x+3 y=18;8 x=7-3 y \). Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}3x+2y=13\\2x+3y=18\\8x=7-3y\end{array}\right.\) - step1: Solve the equation: \(\left\{ \begin{array}{l}x=\frac{13-2y}{3}\\2x+3y=18\\8x=7-3y\end{array}\right.\) - step2: Substitute the value of \(x:\) \(\left\{ \begin{array}{l}2\times \frac{13-2y}{3}+3y=18\\8\times \frac{13-2y}{3}=7-3y\end{array}\right.\) - step3: Simplify: \(\left\{ \begin{array}{l}\frac{2\left(13-2y\right)}{3}+3y=18\\\frac{8\left(13-2y\right)}{3}=7-3y\end{array}\right.\) - step4: Solve the equation: \(\left\{ \begin{array}{l}y=\frac{28}{5}\\\frac{8\left(13-2y\right)}{3}=7-3y\end{array}\right.\) - step5: Substitute the value of \(y:\) \(\frac{8\left(13-2\times \frac{28}{5}\right)}{3}=7-3\times \frac{28}{5}\) - step6: Simplify: \(\frac{24}{5}=-\frac{49}{5}\) - step7: Calculate: \(\left\{ \begin{array}{l}x \in \varnothing \\y \in \varnothing \end{array}\right.\) - step8: Rewrite: \((x, y) \in \varnothing\) Solve the system of equations \( 3 y-4 x=4;2 y=x;8 x-1=-2 y \). Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}3y-4x=4\\2y=x\\8x-1=-2y\end{array}\right.\) - step1: Rewrite the expression: \(\left\{ \begin{array}{l}3y-4x=4\\x=2y\\8x-1=-2y\end{array}\right.\) - step2: Substitute the value of \(x:\) \(\left\{ \begin{array}{l}3y-4\times 2y=4\\8\times 2y-1=-2y\end{array}\right.\) - step3: Simplify: \(\left\{ \begin{array}{l}-5y=4\\16y-1=-2y\end{array}\right.\) - step4: Solve the equation: \(\left\{ \begin{array}{l}y=-\frac{4}{5}\\16y-1=-2y\end{array}\right.\) - step5: Substitute the value of \(y:\) \(16\left(-\frac{4}{5}\right)-1=-2\left(-\frac{4}{5}\right)\) - step6: Simplify: \(-\frac{69}{5}=\frac{8}{5}\) - step7: Calculate: \(\left\{ \begin{array}{l}x \in \varnothing \\y \in \varnothing \end{array}\right.\) - step8: Rewrite: \((x, y) \in \varnothing\) Solve the system of equations \( 8 x=7-3 y;3 x+4 y=3;9 y=-20 x+13;6 x=7-10 y \). Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}8x=7-3y\\3x+4y=3\\9y=-20x+13\\6x=7-10y\end{array}\right.\) - step1: Solve the equation: \(\left\{ \begin{array}{l}x=\frac{7-3y}{8}\\3x+4y=3\\9y=-20x+13\\6x=7-10y\end{array}\right.\) - step2: Substitute the value of \(x:\) \(\left\{ \begin{array}{l}3\times \frac{7-3y}{8}+4y=3\\9y=-20\times \frac{7-3y}{8}+13\\6\times \frac{7-3y}{8}=7-10y\end{array}\right.\) - step3: Simplify: \(\left\{ \begin{array}{l}\frac{3\left(7-3y\right)}{8}+4y=3\\9y=-\frac{5\left(7-3y\right)}{2}+13\\\frac{3\left(7-3y\right)}{4}=7-10y\end{array}\right.\) - step4: Solve the equation: \(\left\{ \begin{array}{l}y=\frac{3}{23}\\9y=-\frac{5\left(7-3y\right)}{2}+13\\\frac{3\left(7-3y\right)}{4}=7-10y\end{array}\right.\) - step5: Substitute the value of \(y:\) \(\left\{ \begin{array}{l}9\times \frac{3}{23}=-\frac{5\left(7-3\times \frac{3}{23}\right)}{2}+13\\\frac{3\left(7-3\times \frac{3}{23}\right)}{4}=7-10\times \frac{3}{23}\end{array}\right.\) - step6: Simplify: \(\left\{ \begin{array}{l}\frac{27}{23}=-\frac{81}{23}\\\frac{114}{23}=\frac{131}{23}\end{array}\right.\) - step7: Calculate: \(\left\{ \begin{array}{l}x \in \varnothing \\y \in \varnothing \end{array}\right.\) - step8: Rewrite: \((x, y) \in \varnothing\) Let's solve the systems of equations step by step based on the provided inputs. ### 1. System: \[ x + y = 10 \] \[ x - y = 2 \] **Solution:** By solving these equations, we find: \[ (x, y) = (6, 4) \] ### 2. System: \[ x + y = 5 \] **Solution:** This is a single equation with infinite solutions. We can express \(y\) in terms of \(x\): \[ y = 5 - x \] ### 3. System: \[ 2x + 3y = 11 \] \[ 4y - 2x = 10 \] **Solution:** By solving these equations, we find: \[ (x, y) = (1, 3) \] ### 4. System: \[ 3x + 2y = 13 \] \[ 2x + 3y = 18 \] \[ 8x = 7 - 3y \] **Solution:** This system has no solution, which can be expressed as: \[ (x, y) \in \varnothing \quad \text{(No solution)} \] ### 5. System: \[ 8x = 7 - 3y \] \[ 3x + 4y = 3 \] \[ 9y = -20x + 13 \] \[ 6x = 7 - 10y \] **Solution:** This system also has no solution: \[ (x, y) \in \varnothing \quad \text{(No solution)} \] ### 6. System: \[ 2y = 3x + 3 \] **Solution:** This is a single equation with infinite solutions. We can express \(y\) in terms of \(x\): \[ y = \frac{3}{2}x + \frac{3}{2} \] ### 7. System: \[ 3y - 4x = 4 \] \[ 2y = x \] \[ 8x - 1 = -2y \] **Solution:** This system has no solution: \[ (x, y) \in \varnothing \quad \text{(No solution)} \] ### 8. System: \[ 4y = -x \] **Solution:** This is a single equation with infinite solutions. We can express \(y\) in terms of \(x\): \[ y = -\frac{1}{4}x \] ### 9. System: \[ 3y + 6 = 6x \] \[ x = 2y - 2 \] **Solution:** By solving these equations, we find: \[ (x, y) = (2, 2) \] ### Summary of Solutions: 1. \( (x, y) = (6, 4) \) 2. Infinite solutions: \( y = 5 - x \) 3. \( (x, y) = (1, 3) \) 4. No solution 5. No solution 6. Infinite solutions: \( y = \frac{3}{2}x + \frac{3}{2} \) 7. No solution 8. Infinite solutions: \( y = -\frac{1}{4}x \) 9. \( (x, y) = (2, 2) \)