5. \( 2 x-2 y=0 \) 6. \( 2 x+3 y=12 \) and \( x-y-1=0 \) 7. \( x+2 y=4 \) and \( 2 x-y-3=0 \) 8. \( 2 y-5 x=7 \) and \( 4 x+3 y+24= \)

Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}2x+3y=12\\x-y-1=0\end{array}\right.\) - step1: Solve the equation: \(\left\{ \begin{array}{l}2x+3y=12\\x=y+1\end{array}\right.\) - step2: Substitute the value of \(x:\) \(2\left(y+1\right)+3y=12\) - step3: Simplify: \(5y+2=12\) - step4: Move the constant to the right side: \(5y=12-2\) - step5: Subtract the numbers: \(5y=10\) - step6: Divide both sides: \(\frac{5y}{5}=\frac{10}{5}\) - step7: Divide the numbers: \(y=2\) - step8: Substitute the value of \(y:\) \(x=2+1\) - step9: Calculate: \(x=3\) - step10: Calculate: \(\left\{ \begin{array}{l}x=3\\y=2\end{array}\right.\) - step11: Check the solution: \(\left\{ \begin{array}{l}x=3\\y=2\end{array}\right.\) - step12: Rewrite: \(\left(x,y\right) = \left(3,2\right)\) Solve the system of equations \( 2 y-5 x=7; 4 x+3 y=0 \). Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}2y-5x=7\\4x+3y=0\end{array}\right.\) - step1: Solve the equation: \(\left\{ \begin{array}{l}x=\frac{-7+2y}{5}\\4x+3y=0\end{array}\right.\) - step2: Substitute the value of \(x:\) \(4\times \frac{-7+2y}{5}+3y=0\) - step3: Simplify: \(\frac{4\left(-7+2y\right)}{5}+3y=0\) - step4: Multiply both sides of the equation by LCD: \(\left(\frac{4\left(-7+2y\right)}{5}+3y\right)\times 5=0\times 5\) - step5: Simplify the equation: \(-28+23y=0\) - step6: Move the constant to the right side: \(23y=0+28\) - step7: Remove 0: \(23y=28\) - step8: Divide both sides: \(\frac{23y}{23}=\frac{28}{23}\) - step9: Divide the numbers: \(y=\frac{28}{23}\) - step10: Substitute the value of \(y:\) \(x=\frac{-7+2\times \frac{28}{23}}{5}\) - step11: Calculate: \(x=-\frac{21}{23}\) - step12: Calculate: \(\left\{ \begin{array}{l}x=-\frac{21}{23}\\y=\frac{28}{23}\end{array}\right.\) - step13: Check the solution: \(\left\{ \begin{array}{l}x=-\frac{21}{23}\\y=\frac{28}{23}\end{array}\right.\) - step14: Rewrite: \(\left(x,y\right) = \left(-\frac{21}{23},\frac{28}{23}\right)\) Solve the system of equations \( x+2 y=4; 2 x-y-3=0 \). Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}x+2y=4\\2x-y-3=0\end{array}\right.\) - step1: Solve the equation: \(\left\{ \begin{array}{l}x=4-2y\\2x-y-3=0\end{array}\right.\) - step2: Substitute the value of \(x:\) \(2\left(4-2y\right)-y-3=0\) - step3: Simplify: \(5-5y=0\) - step4: Move the constant to the right side: \(-5y=0-5\) - step5: Remove 0: \(-5y=-5\) - step6: Change the signs: \(5y=5\) - step7: Divide both sides: \(\frac{5y}{5}=\frac{5}{5}\) - step8: Divide the numbers: \(y=1\) - step9: Substitute the value of \(y:\) \(x=4-2\times 1\) - step10: Substitute back: \(x=4-2\) - step11: Calculate: \(x=2\) - step12: Calculate: \(\left\{ \begin{array}{l}x=2\\y=1\end{array}\right.\) - step13: Check the solution: \(\left\{ \begin{array}{l}x=2\\y=1\end{array}\right.\) - step14: Rewrite: \(\left(x,y\right) = \left(2,1\right)\) Solve the equation \( 2 x-2 y=0 \). Solve the equation by following steps: - step0: Solve for \(y\): \(2x-2y=0\) - step1: Move the expression to the right side: \(-2y=0-2x\) - step2: Remove 0: \(-2y=-2x\) - step3: Change the signs: \(2y=2x\) - step4: Divide both sides: \(\frac{2y}{2}=\frac{2x}{2}\) - step5: Divide the numbers: \(y=x\) Here are the solutions for each of the equations and systems of equations provided: 5. For the equation \( 2x - 2y = 0 \): - Rearranging gives \( 2y = 2x \) or \( y = x \). - **Solution**: \( y = x \). 6. For the system of equations: - \( 2x + 3y = 12 \) - \( x - y - 1 = 0 \) (which can be rewritten as \( x = y + 1 \)) - Substituting \( x = y + 1 \) into the first equation: \[ 2(y + 1) + 3y = 12 \implies 2y + 2 + 3y = 12 \implies 5y + 2 = 12 \implies 5y = 10 \implies y = 2 \] - Then substituting \( y = 2 \) back to find \( x \): \[ x = 2 + 1 = 3 \] - **Solution**: \( (x, y) = (3, 2) \). 7. For the system of equations: - \( x + 2y = 4 \) - \( 2x - y - 3 = 0 \) (which can be rewritten as \( y = 2x - 3 \)) - Substituting \( y = 2x - 3 \) into the first equation: \[ x + 2(2x - 3) = 4 \implies x + 4x - 6 = 4 \implies 5x - 6 = 4 \implies 5x = 10 \implies x = 2 \] - Then substituting \( x = 2 \) back to find \( y \): \[ y = 2(2) - 3 = 4 - 3 = 1 \] - **Solution**: \( (x, y) = (2, 1) \). 8. For the system of equations: - \( 2y - 5x = 7 \) - \( 4x + 3y = -24 \) (correcting the equation to equal \(-24\)) - Rearranging the first equation gives \( 2y = 5x + 7 \) or \( y = \frac{5}{2}x + \frac{7}{2} \). - Substituting \( y \) into the second equation: \[ 4x + 3\left(\frac{5}{2}x + \frac{7}{2}\right) = -24 \implies 4x + \frac{15}{2}x + \frac{21}{2} = -24 \] \[ \implies \left(4 + \frac{15}{2}\right)x + \frac{21}{2} = -24 \implies \frac{8}{2} + \frac{15}{2} = \frac{23}{2} \implies \frac{23}{2}x + \frac{21}{2} = -24 \] \[ \implies \frac{23}{2}x = -24 - \frac{21}{2} \implies \frac{23}{2}x = -\frac{48}{2} - \frac{21}{2} \implies \frac{23}{2}x = -\frac{69}{2} \implies x = -\frac{69}{23} = -3 \] - Then substituting \( x = -3 \) back to find \( y \): \[ 2y - 5(-3) = 7 \implies 2y + 15 = 7 \implies 2y = 7 - 15 \implies 2y = -8 \implies y = -4 \] - **Solution**: \( (x, y) = \left(-\frac{21}{23}, \frac{28}{23}\right) \). In summary: - 5: \( y = x \) - 6: \( (x, y) = (3, 2) \) - 7: \( (x, y) = (2, 1) \) - 8: \( (x, y) = \left(-\frac{21}{23}, \frac{28}{23}\right) \)