Question

solving systems of linear equations: graphing
how many solutions does this linear system have?
y = 2x - 5
-8x - 4y = -20
one solution: (-2.5, 0)
one solution: (2.5, 0)
infinite number of solutions
no solution

Explanation:

Step1: Simplify the second equation

Start with \(-8x - 4y = -20\). Divide every term by \(-4\) to simplify.
\(\frac{-8x}{-4}+\frac{-4y}{-4}=\frac{-20}{-4}\)
Which simplifies to \(2x + y = 5\). Now, solve for \(y\): \(y=-2x + 5\)? Wait, no, wait. Wait, original first equation is \(y = 2x - 5\). Wait, let's re - do the simplification. Wait, \(-8x-4y=-20\), divide both sides by \(-4\): \(\frac{-8x}{-4}+\frac{-4y}{-4}=\frac{-20}{-4}\) gives \(2x + y = 5\), so \(y=-2x + 5\)? Wait, no, that's a mistake. Wait, no: \(-8x-4y=-20\), let's solve for \(y\) directly. Add \(8x\) to both sides: \(-4y=8x - 20\), then divide by \(-4\): \(y=-2x + 5\)? Wait, but the first equation is \(y = 2x - 5\). Wait, no, maybe I made a mistake in simplifying. Wait, let's check again. The second equation: \(-8x-4y=-20\). Let's divide all terms by \(-4\): \(\frac{-8x}{-4}=2x\), \(\frac{-4y}{-4}=y\), \(\frac{-20}{-4}=5\). So the equation becomes \(2x + y = 5\), so \(y=-2x + 5\)? Wait, no, that's not right. Wait, no, if we have \(-8x-4y=-20\), let's solve for \(y\):

\(-4y=8x - 20\)

\(y=\frac{8x - 20}{-4}=\frac{8x}{-4}-\frac{20}{-4}=-2x + 5\). Wait, but the first equation is \(y = 2x - 5\). Wait, that can't be. Wait, maybe I messed up the sign. Wait, let's take the second equation: \(-8x-4y=-20\). Let's add \(8x\) to both sides: \(-4y=8x - 20\). Then divide by \(-4\): \(y=-2x + 5\). The first equation is \(y = 2x - 5\). Wait, these are two lines with slopes \(2\) and \(-2\)? Wait, no, that's not possible. Wait, maybe I made a mistake in simplifying. Wait, let's try another approach. Let's substitute \(y = 2x - 5\) into the second equation.

Second equation: \(-8x-4y=-20\)

Substitute \(y = 2x - 5\) into it:

\(-8x-4(2x - 5)=-20\)

Step2: Expand and solve

Expand the left - hand side: \(-8x-8x + 20=-20\)

Combine like terms: \(-16x+20=-20\)

Subtract 20 from both sides: \(-16x=-40\)

Divide both sides by \(-16\): \(x=\frac{-40}{-16}=\frac{5}{2}=2.5\)

Now substitute \(x = 2.5\) into \(y = 2x - 5\):

\(y=2(2.5)-5=5 - 5 = 0\)

Wait, but earlier when I simplified the second equation, I got a wrong slope. Wait, let's re - simplify the second equation correctly.

Second equation: \(-8x-4y=-20\)

Let's divide all terms by \(-4\):

\(\frac{-8x}{-4}=2x\), \(\frac{-4y}{-4}=y\), \(\frac{-20}{-4}=5\)

So the equation is \(2x + y = 5\), which can be written as \(y=-2x + 5\)? Wait, no, that's incorrect. Wait, \(2x + y = 5\) implies \(y=-2x + 5\)? Wait, no, \(2x + y = 5\) => \(y=-2x + 5\). But when we substitute \(y = 2x - 5\) into the second equation, we get a valid solution. Wait, there must be a mistake in the slope calculation. Wait, no, if we have \(y = 2x - 5\) and \(y=-2x + 5\), these are two lines with slopes \(2\) and \(-2\), which are not parallel (since slopes are not equal) and not the same line (since slopes are different). So they should intersect at one point. But when we solved, we got \(x = 2.5\) and \(y = 0\). Wait, let's check if \((2.5,0)\) satisfies the second equation.

Second equation: \(-8x-4y=-20\)

Substitute \(x = 2.5\) and \(y = 0\):

Left - hand side: \(-8(2.5)-4(0)=-20 - 0=-20\), which is equal to the right - hand side. And it satisfies the first equation: \(y = 2(2.5)-5=5 - 5 = 0\). Wait, but earlier when I simplified the second equation, I must have made a mistake. Wait, no, \(2x + y = 5\) when \(x = 2.5\) and \(y = 0\): \(2(2.5)+0 = 5\), which is true. And \(y = 2x - 5\) when \(x = 2.5\), \(y = 0\) is also true. Wait, so my initial simplification of the second equation to \(y=-2x + 5\) was wrong? Wait, no, \(2x + y = 5\) => \(y=-2x + 5\). But when \(x = 2.5\), \(y=-2(2.5)+5=-5 + 5 = 0\), which is the same as before. Oh! I see, I made a mistake in the sign when I thought the slopes were different. Wait, \(y = 2x - 5\) has a slope of \(2\), and \(y=-2x + 5\) has a slope of \(-2\)? Wait, no, that's not possible. Wait, no, \(2x + y = 5\) is \(y=-2x + 5\), and \(y = 2x - 5\). These two lines have slopes \(2\) and \(-2\), which are not equal and not negative reciprocals (they are not perpendicular, just intersecting). So they intersect at one point \((2.5,0)\).

Wait, but let's check the two equations again. The first equation: \(y = 2x - 5\). The second equation: \(-8x-4y=-20\). Let's rewrite the second equation in slope - intercept form correctly.

Starting with \(-8x-4y=-20\)

Add \(8x\) to both sides: \(-4y=8x - 20\)

Divide both sides by \(-4\): \(y=-2x + 5\)

Now, the first equation is \(y = 2x - 5\), the second is \(y=-2x + 5\). Wait, these are two lines with slopes \(2\) and \(-2\), so they are not parallel (since slopes are not equal) and not the same line (since slopes and y - intercepts are different). So they intersect at exactly one point. We found that point by substitution: when \(x = 2.5\), \(y = 0\). So the system has one solution \((2.5,0)\)

Answer:

one solution: \((2.5,0)\)