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Solve graphically \( \begin{array}{l}x+4 y=15 \\ x+2 y=7\end{array} \)

Ask by Davies Davies. in the United States
Feb 19,2025

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Tutor-Verified Answer

Answer

The solution is \( x=-1 \) and \( y=4 \).

Solution

Solve the system of equations \( x+4y=15;x+2y=7 \). Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}x+4y=15\\x+2y=7\end{array}\right.\) - step1: Solve the equation: \(\left\{ \begin{array}{l}x=15-4y\\x+2y=7\end{array}\right.\) - step2: Substitute the value of \(x:\) \(15-4y+2y=7\) - step3: Add the terms: \(15-2y=7\) - step4: Move the constant to the right side: \(-2y=7-15\) - 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 \(y:\) \(x=15-4\times 4\) - step10: Calculate: \(x=-1\) - step11: Calculate: \(\left\{ \begin{array}{l}x=-1\\y=4\end{array}\right.\) - step12: Check the solution: \(\left\{ \begin{array}{l}x=-1\\y=4\end{array}\right.\) - step13: Rewrite: \(\left(x,y\right) = \left(-1,4\right)\) The solution to the system of equations \(x+4y=15\) and \(x+2y=7\) is \(x=-1\) and \(y=4\).

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Bonus Knowledge

To solve the system of equations graphically, start by rearranging each equation into slope-intercept form (y = mx + b). For the first equation \( x + 4y = 15 \), isolate y: \[ 4y = 15 - x \implies y = -\frac{1}{4}x + \frac{15}{4} \] For the second equation \( x + 2y = 7 \): \[ 2y = 7 - x \implies y = -\frac{1}{2}x + \frac{7}{2} \] Next, plot both lines on a graph using the y-intercepts and slopes: - The first line crosses the y-axis at \( \frac{15}{4} \) and has a slope of \( -\frac{1}{4} \). - The second line crosses the y-axis at \( \frac{7}{2} \) and has a slope of \( -\frac{1}{2} \). By finding the intersection point of the two lines, you’ll obtain the solution to the system. The intersection coordinates are the values of \( x \) and \( y \) that satisfy both equations simultaneously. This intersection can be calculated algebraically and validated through graphical representation. Try plotting the lines to find the point where they intersect visually!

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