Sirnultaneous Equations Evercise 7: 1) Solve for \( x \) and \( y \) in each of the following equations: a) \( y+6 x=4 \) and f) \( y-x=-6 \) and \[ 2 y=5 x^{2}-22 x+8 \] (b) \( 2 x-y=8 \) and \[ y=x^{2}+4 x-23 \] g. \( y-x=5 \) and \[ 2 x^{2}+5 x y=-2 y^{2} \] \( x y-3 y=8 x-20 \) c) \( y+x=6 \) and h) \[ \begin{array}{l} 2^{x+y}=256 \text { and } \\ x y+3 x+2 y=36 \end{array} \] d) \( 2 x-3 y=2 \) and i) \( 5^{x-2 y-1}=1 \) and \( x^{2}-2 y^{2}=8 \) \[ x^{2}+2 x y-2 x-4 y=0 \] e) \( y-2=2(x-1)^{2} \) and j) \( \cdot(x+5)^{2}+(y-3)^{2}=49 \) and \( y-x=2 \) \( y+x=-9 \)

Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}y+x=6\\2^{x+y}=256\end{array}\right.\) - step1: Solve the equation: \(\left\{ \begin{array}{l}x=6-y\\2^{x+y}=256\end{array}\right.\) - step2: Substitute the value of \(x:\) \(2^{6-y+y}=256\) - step3: Simplify: \(64=256\) - step4: Calculate: \(\left\{ \begin{array}{l}x \in \varnothing \\y \in \varnothing \end{array}\right.\) - step5: Rewrite: \((x, y) \in \varnothing\) Solve the system of equations \( y-x=2; y+x=-9 \). Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}y-x=2\\y+x=-9\end{array}\right.\) - step1: Solve the equation: \(\left\{ \begin{array}{l}x=-2+y\\y+x=-9\end{array}\right.\) - step2: Substitute the value of \(x:\) \(y-2+y=-9\) - step3: Add the terms: \(2y-2=-9\) - step4: Move the constant to the right side: \(2y=-9+2\) - step5: Add the numbers: \(2y=-7\) - step6: Divide both sides: \(\frac{2y}{2}=\frac{-7}{2}\) - step7: Divide the numbers: \(y=-\frac{7}{2}\) - step8: Substitute the value of \(y:\) \(x=-2-\frac{7}{2}\) - step9: Calculate: \(x=-\frac{11}{2}\) - step10: Calculate: \(\left\{ \begin{array}{l}x=-\frac{11}{2}\\y=-\frac{7}{2}\end{array}\right.\) - step11: Check the solution: \(\left\{ \begin{array}{l}x=-\frac{11}{2}\\y=-\frac{7}{2}\end{array}\right.\) - step12: Rewrite: \(\left(x,y\right) = \left(-\frac{11}{2},-\frac{7}{2}\right)\) Solve the system of equations \( 5^{x-2 y-1}=1; x^{2}-2 y^{2}=8 \). Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}5^{x-2y-1}=1\\x^{2}-2y^{2}=8\end{array}\right.\) - step1: Solve the equation: \(\left\{ \begin{array}{l}x=2y+1\\x^{2}-2y^{2}=8\end{array}\right.\) - step2: Substitute the value of \(x:\) \(\left(2y+1\right)^{2}-2y^{2}=8\) - step3: Simplify: \(2y^{2}+4y+1=8\) - step4: Move the expression to the left side: \(2y^{2}+4y+1-8=0\) - step5: Subtract the numbers: \(2y^{2}+4y-7=0\) - step6: Solve using the quadratic formula: \(y=\frac{-4\pm \sqrt{4^{2}-4\times 2\left(-7\right)}}{2\times 2}\) - step7: Simplify the expression: \(y=\frac{-4\pm \sqrt{4^{2}-4\times 2\left(-7\right)}}{4}\) - step8: Simplify the expression: \(y=\frac{-4\pm \sqrt{72}}{4}\) - step9: Simplify the expression: \(y=\frac{-4\pm 6\sqrt{2}}{4}\) - step10: Separate into possible cases: \(\begin{align}&y=\frac{-4+6\sqrt{2}}{4}\\&y=\frac{-4-6\sqrt{2}}{4}\end{align}\) - step11: Simplify the expression: \(\begin{align}&y=\frac{-2+3\sqrt{2}}{2}\\&y=\frac{-4-6\sqrt{2}}{4}\end{align}\) - step12: Simplify the expression: \(\begin{align}&y=\frac{-2+3\sqrt{2}}{2}\\&y=-\frac{2+3\sqrt{2}}{2}\end{align}\) - step13: Evaluate the logic: \(y=\frac{-2+3\sqrt{2}}{2}\cup y=-\frac{2+3\sqrt{2}}{2}\) - step14: Rearrange the terms: \(\left\{ \begin{array}{l}x=2y+1\\y=\frac{-2+3\sqrt{2}}{2}\end{array}\right.\cup \left\{ \begin{array}{l}x=2y+1\\y=-\frac{2+3\sqrt{2}}{2}\end{array}\right.\) - step15: Calculate: \(\left\{ \begin{array}{l}x=-1+3\sqrt{2}\\y=\frac{-2+3\sqrt{2}}{2}\end{array}\right.\cup \left\{ \begin{array}{l}x=-1-3\sqrt{2}\\y=-\frac{2+3\sqrt{2}}{2}\end{array}\right.\) - step16: Check the solution: \(\left\{ \begin{array}{l}x=-1+3\sqrt{2}\\y=\frac{-2+3\sqrt{2}}{2}\end{array}\right.\cup \left\{ \begin{array}{l}x=-1-3\sqrt{2}\\y=-\frac{2+3\sqrt{2}}{2}\end{array}\right.\) - step17: Rewrite: \(\left(x,y\right) = \left(-1+3\sqrt{2},\frac{-2+3\sqrt{2}}{2}\right)\cup \left(x,y\right) = \left(-1-3\sqrt{2},-\frac{2+3\sqrt{2}}{2}\right)\) Solve the system of equations \( y+6 x=4; y-x=-6 \). Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}y+6x=4\\y-x=-6\end{array}\right.\) - step1: Solve the equation: \(\left\{ \begin{array}{l}y=4-6x\\y-x=-6\end{array}\right.\) - step2: Substitute the value of \(y:\) \(4-6x-x=-6\) - step3: Subtract the terms: \(4-7x=-6\) - step4: Move the constant to the right side: \(-7x=-6-4\) - step5: Subtract the numbers: \(-7x=-10\) - step6: Change the signs: \(7x=10\) - step7: Divide both sides: \(\frac{7x}{7}=\frac{10}{7}\) - step8: Divide the numbers: \(x=\frac{10}{7}\) - step9: Substitute the value of \(x:\) \(y=4-6\times \frac{10}{7}\) - step10: Calculate: \(y=-\frac{32}{7}\) - step11: Calculate: \(\left\{ \begin{array}{l}x=\frac{10}{7}\\y=-\frac{32}{7}\end{array}\right.\) - step12: Check the solution: \(\left\{ \begin{array}{l}x=\frac{10}{7}\\y=-\frac{32}{7}\end{array}\right.\) - step13: Rewrite: \(\left(x,y\right) = \left(\frac{10}{7},-\frac{32}{7}\right)\) Solve the system of equations \( 2 x-y=8; y=x^{2}+4 x-23 \). Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}2x-y=8\\y=x^{2}+4x-23\end{array}\right.\) - step1: Substitute the value of \(y:\) \(2x-\left(x^{2}+4x-23\right)=8\) - step2: Simplify: \(-2x-x^{2}+23=8\) - step3: Move the expression to the left side: \(-2x-x^{2}+23-8=0\) - step4: Subtract the numbers: \(-2x-x^{2}+15=0\) - step5: Factor the expression: \(\left(-x+3\right)\left(x+5\right)=0\) - step6: Separate into possible cases: \(\begin{align}&-x+3=0\\&x+5=0\end{align}\) - step7: Solve the equation: \(\begin{align}&x=3\\&x=-5\end{align}\) - step8: Calculate: \(x=3\cup x=-5\) - step9: Rearrange the terms: \(\left\{ \begin{array}{l}x=3\\y=x^{2}+4x-23\end{array}\right.\cup \left\{ \begin{array}{l}x=-5\\y=x^{2}+4x-23\end{array}\right.\) - step10: Calculate: \(\left\{ \begin{array}{l}x=3\\y=-2\end{array}\right.\cup \left\{ \begin{array}{l}x=-5\\y=-18\end{array}\right.\) - step11: Calculate: \(\left\{ \begin{array}{l}x=-5\\y=-18\end{array}\right.\cup \left\{ \begin{array}{l}x=3\\y=-2\end{array}\right.\) - step12: Check the solution: \(\left\{ \begin{array}{l}x=-5\\y=-18\end{array}\right.\cup \left\{ \begin{array}{l}x=3\\y=-2\end{array}\right.\) - step13: Rewrite: \(\left(x,y\right) = \left(-5,-18\right)\cup \left(x,y\right) = \left(3,-2\right)\) Solve the system of equations \( y-2=2(x-1)^{2}; (x+5)^{2}+(y-3)^{2}=49 \). Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}y-2=2\left(x-1\right)^{2}\\\left(x+5\right)^{2}+\left(y-3\right)^{2}=49\end{array}\right.\) - step1: Solve the equation: \(\left\{ \begin{array}{l}y=2x^{2}-4x+4\\\left(x+5\right)^{2}+\left(y-3\right)^{2}=49\end{array}\right.\) - step2: Substitute the value of \(y:\) \(\left(x+5\right)^{2}+\left(2x^{2}-4x+4-3\right)^{2}=49\) - step3: Simplify: \(21x^{2}+2x+26+4x^{4}-16x^{3}=49\) - step4: Move the expression to the left side: \(21x^{2}+2x+26+4x^{4}-16x^{3}-49=0\) - step5: Subtract the numbers: \(21x^{2}+2x-23+4x^{4}-16x^{3}=0\) - step6: Calculate: \(x\approx 1.952466\cup x\approx -0.818491\) - step7: Rearrange the terms: \(\left\{ \begin{array}{l}x\approx 1.952466\\y=2x^{2}-4x+4\end{array}\right.\cup \left\{ \begin{array}{l}x\approx -0.818491\\y=2x^{2}-4x+4\end{array}\right.\) - step8: Calculate: \(\left\{ \begin{array}{l}x\approx 1.952466\\y\approx 3.814382\end{array}\right.\cup \left\{ \begin{array}{l}x\approx -0.818491\\y\approx 8.613821\end{array}\right.\) - step9: Calculate: \(\left\{ \begin{array}{l}x\approx -0.818491\\y\approx 8.613821\end{array}\right.\cup \left\{ \begin{array}{l}x\approx 1.952466\\y\approx 3.814382\end{array}\right.\) - step10: Check the solution: \(\left\{ \begin{array}{l}x\approx -0.818491\\y\approx 8.613821\end{array}\right.\cup \left\{ \begin{array}{l}x\approx 1.952466\\y\approx 3.814382\end{array}\right.\) - step11: Rewrite: \(\left(x,y\right)\approx \left(-0.818491,8.613821\right)\cup \left(x,y\right)\approx \left(1.952466,3.814382\right)\) Solve the system of equations \( 2 x-3 y=2; x^{2}+2 x y-2 x-4 y=0 \). Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}2x-3y=2\\x^{2}+2xy-2x-4y=0\end{array}\right.\) - step1: Solve the equation: \(\left\{ \begin{array}{l}x=\frac{2+3y}{2}\\x^{2}+2xy-2x-4y=0\end{array}\right.\) - step2: Substitute the value of \(x:\) \(\left(\frac{2+3y}{2}\right)^{2}+2\times \frac{2+3y}{2}\times y-2\times \frac{2+3y}{2}-4y=0\) - step3: Simplify: \(-1-2y+\frac{21}{4}y^{2}=0\) - step4: Factor the expression: \(\frac{1}{4}\left(-2+3y\right)\left(2+7y\right)=0\) - step5: Divide the terms: \(\left(-2+3y\right)\left(2+7y\right)=0\) - step6: Separate into possible cases: \(\begin{align}&-2+3y=0\\&2+7y=0\end{align}\) - step7: Solve the equation: \(\begin{align}&y=\frac{2}{3}\\&y=-\frac{2}{7}\end{align}\) - step8: Calculate: \(y=\frac{2}{3}\cup y=-\frac{2}{7}\) - step9: Rearrange the terms: \(\left\{ \begin{array}{l}x=\frac{2+3y}{2}\\y=\frac{2}{3}\end{array}\right.\cup \left\{ \begin{array}{l}x=\frac{2+3y}{2}\\y=-\frac{2}{7}\end{array}\right.\) - step10: Calculate: \(\left\{ \begin{array}{l}x=2\\y=\frac{2}{3}\end{array}\right.\cup \left\{ \begin{array}{l}x=\frac{4}{7}\\y=-\frac{2}{7}\end{array}\right.\) - step11: Check the solution: \(\left\{ \begin{array}{l}x=2\\y=\frac{2}{3}\end{array}\right.\cup \left\{ \begin{array}{l}x=\frac{4}{7}\\y=-\frac{2}{7}\end{array}\right.\) - step12: Rewrite: \(\left(x,y\right) = \left(2,\frac{2}{3}\right)\cup \left(x,y\right) = \left(\frac{4}{7},-\frac{2}{7}\right)\) Solve the system of equations \( y-x=5; 2 x^{2}+5 x y=-2 y^{2} \). Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}y-x=5\\2x^{2}+5xy=-2y^{2}\end{array}\right.\) - step1: Solve the equation: \(\left\{ \begin{array}{l}x=-5+y\\2x^{2}+5xy=-2y^{2}\end{array}\right.\) - step2: Substitute the value of \(x:\) \(2\left(-5+y\right)^{2}+5\left(-5+y\right)y=-2y^{2}\) - step3: Simplify: \(50-45y+7y^{2}=-2y^{2}\) - step4: Move the expression to the left side: \(50-45y+7y^{2}-\left(-2y^{2}\right)=0\) - step5: Calculate: \(50-45y+9y^{2}=0\) - step6: Factor the expression: \(\left(10-3y\right)\left(5-3y\right)=0\) - step7: Separate into possible cases: \(\begin{align}&10-3y=0\\&5-3y=0\end{align}\) - step8: Solve the equation: \(\begin{align}&y=\frac{10}{3}\\&y=\frac{5}{3}\end{align}\) - step9: Calculate: \(y=\frac{10}{3}\cup y=\frac{5}{3}\) - step10: Rearrange the terms: \(\left\{ \begin{array}{l}x=-5+y\\y=\frac{10}{3}\end{array}\right.\cup \left\{ \begin{array}{l}x=-5+y\\y=\frac{5}{3}\end{array}\right.\) - step11: Calculate: \(\left\{ \begin{array}{l}x=-\frac{5}{3}\\y=\frac{10}{3}\end{array}\right.\cup \left\{ \begin{array}{l}x=-\frac{10}{3}\\y=\frac{5}{3}\end{array}\right.\) - step12: Calculate: \(\left\{ \begin{array}{l}x=-\frac{10}{3}\\y=\frac{5}{3}\end{array}\right.\cup \left\{ \begin{array}{l}x=-\frac{5}{3}\\y=\frac{10}{3}\end{array}\right.\) - step13: Check the solution: \(\left\{ \begin{array}{l}x=-\frac{10}{3}\\y=\frac{5}{3}\end{array}\right.\cup \left\{ \begin{array}{l}x=-\frac{5}{3}\\y=\frac{10}{3}\end{array}\right.\) - step14: Rewrite: \(\left(x,y\right) = \left(-\frac{10}{3},\frac{5}{3}\right)\cup \left(x,y\right) = \left(-\frac{5}{3},\frac{10}{3}\right)\) Here are the solutions for each of the simultaneous equations provided: ### a) Equations: 1. \( y + 6x = 4 \) 2. \( y - x = -6 \) **Solution:** \[ (x,y) = \left(\frac{10}{7}, -\frac{32}{7}\right) \] ### b) Equations: 1. \( 2x - y = 8 \) 2. \( y = x^2 + 4x - 23 \) **Solution:** \[ (x,y) = (-5,-18) \cup (x,y) = (3,-2) \] ### c) Equations: 1. \( y - x = 5 \) 2. \( 2x^2 + 5xy = -2y^2 \) **Solution:** \[ (x,y) = \left(-\frac{10}{3}, \frac{5}{3}\right) \cup (x,y) = \left(-\frac{5}{3}, \frac{10}{3}\right) \] ### d) Equations: 1. \( y + x = 6 \) 2. \( 2^{x+y} = 256 \) **Solution:** \[ (x, y) \in \varnothing \quad \text{(No solution)} \] ### e) Equations: 1. \( 2x - 3y = 2 \) 2. \( x^2 + 2xy - 2x - 4y = 0 \) **Solution:** \[ (x,y) = (2,\frac{2}{3}) \cup (x,y) = \left(\frac{4}{7}, -\frac{2}{7}\right) \] ### f) Equations: 1. \( y - 2 = 2(x - 1)^2 \) 2. \( (x + 5)^2 + (y - 3)^2 = 49 \) **Solution:** \[ (x,y) \approx (-0.818491, 8.613821) \cup (x,y) \approx (1.952466, 3.814382) \] ### g) Equations: 1. \( y - x = 2 \) 2. \( y + x = -9 \) **Solution:** \[ (x,y) = \left(-\frac{11}{2}, -\frac{7}{2}\right) \] ### h) Equations: 1. \( 5^{x - 2y - 1} = 1 \) 2. \( x^2 - 2y^2 = 8 \) **Solution:** \[ (x,y) = \left(-1 + 3\sqrt{2}, \frac{-2 + 3\sqrt{2}}{2}\right) \cup (x,y) = \left(-1 - 3\sqrt{2}, -\frac{2 + 3\sqrt{2}}{2}\right) \] These solutions provide the values of \( x \) and \( y \) for each pair of equations. 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