Problem

Solve the following system of equations, and find the slope for each equation: \[\begin{matrix} 2x + 3y = 6 \ 5x - 4y = -2 \end{matrix}\]

Answer

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Answer

Now, we find the slope of each equation by rearranging them into slope-intercept form. For the first equation, we have \[y = -\frac{2}{3}x + 2\], so the slope is \[-\frac{2}{3}\]. For the second equation, we have \[y = \frac{5}{4}x + \frac{1}{2}\], so the slope is \[\frac{5}{4}\].

Steps

Step 1 :First, we solve the system of equations. We start by finding the value of x. Multiply the first equation by 4 and the second one by 3 to make the coefficients of y the same, then we subtract the second equation from the first one. We get: \[8x + 12y = 24\] and \[15x - 12y = -6\]. Adding these two equations, we get \[23x = 18\], so \[x = \frac{18}{23}\].

Step 2 :Next, we substitute \(x = \frac{18}{23}\) into the first equation to solve for y. We get \[2*\frac{18}{23} + 3y = 6\], which simplifies to \[\frac{36}{23} + 3y = 6\]. Subtracting \[\frac{36}{23}\] from both sides, we get \[3y = \frac{102}{23}\], so \[y = \frac{34}{23}\].

Step 3 :Now, we find the slope of each equation by rearranging them into slope-intercept form. For the first equation, we have \[y = -\frac{2}{3}x + 2\], so the slope is \[-\frac{2}{3}\]. For the second equation, we have \[y = \frac{5}{4}x + \frac{1}{2}\], so the slope is \[\frac{5}{4}\].

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