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Solve the system of equations \( 2 x+3 y=6 \) and \( -3 x-5 y=-12 \) by combining the equations. \[ (2 x+3 y=6) \] \[ (-3 x-5 y=-12) \] \[ \begin{aligned} 2 x+3 y & =6 \\ -3 x-5 y & =-12 \\ \hline \text { o } x+y & =\square \end{aligned} \]

Ask by Burgess Gordon. in the United States
Jan 22,2025

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Tutor-Verified Answer

Answer

The solution is \( x = -6 \) and \( y = 6 \).

Solution

Solve the system of equations \( 2x+3y=6;-3x-5y=-12 \). Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}2x+3y=6\\-3x-5y=-12\end{array}\right.\) - step1: Solve the equation: \(\left\{ \begin{array}{l}x=\frac{6-3y}{2}\\-3x-5y=-12\end{array}\right.\) - step2: Substitute the value of \(x:\) \(-3\times \frac{6-3y}{2}-5y=-12\) - step3: Simplify: \(-\frac{3\left(6-3y\right)}{2}-5y=-12\) - step4: Multiply both sides of the equation by LCD: \(\left(-\frac{3\left(6-3y\right)}{2}-5y\right)\times 2=-12\times 2\) - step5: Simplify the equation: \(-18-y=-24\) - step6: Move the constant to the right side: \(-y=-24+18\) - step7: Add the numbers: \(-y=-6\) - step8: Change the signs: \(y=6\) - step9: Substitute the value of \(y:\) \(x=\frac{6-3\times 6}{2}\) - step10: Simplify: \(x=-6\) - step11: Calculate: \(\left\{ \begin{array}{l}x=-6\\y=6\end{array}\right.\) - step12: Check the solution: \(\left\{ \begin{array}{l}x=-6\\y=6\end{array}\right.\) - step13: Rewrite: \(\left(x,y\right) = \left(-6,6\right)\) The solution to the system of equations \(2x+3y=6\) and \(-3x-5y=-12\) is \(x=-6\) and \(y=6\).

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Beyond the Answer

To solve the system of equations \( 2x + 3y = 6 \) and \( -3x - 5y = -12 \) by combining them, we can first multiply the first equation by 3 and the second by 2 to make the coefficients of \( x \) equal. 1. Multiply the first equation \( 2x + 3y = 6 \) by 3: \[ 6x + 9y = 18 \] 2. Multiply the second equation \( -3x - 5y = -12 \) by 2: \[ -6x - 10y = -24 \] Now, we can add the two modified equations: \[ (6x + 9y) + (-6x - 10y) = 18 + (-24) \] This simplifies to: \[ (6x - 6x) + (9y - 10y) = 18 - 24 \] \[ 0 - y = -6 \] \[ y = 6 \] Now that we have \( y \), we can substitute \( y = 6 \) back into the original first equation to find \( x \): \[ 2x + 3(6) = 6 \] \[ 2x + 18 = 6 \] \[ 2x = 6 - 18 \] \[ 2x = -12 \] \[ x = -6 \] Thus, the solution to the system is: \[ \begin{aligned} x &= -6 \\ y &= 6 \end{aligned} \] Substituting back into the provided structure gives: \[ \begin{aligned} 2x + 3y &= 6 \\ -3x - 5y &= -12 \\ \hline \text{So } x + y &= -6 + 6 = 0 \end{aligned} \] In conclusion: \[ x + y = 0 \text{ (which means }\square = 0\text{)} \]

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