Question

5x + 2y = 8
a) graph of a line
b) graph of a line

Explanation:

Step 1: Rewrite the equation in slope - intercept form

The slope - intercept form of a linear equation is \(y=mx + b\), where \(m\) is the slope and \(b\) is the y - intercept.
Starting with \(5x + 2y=8\), we solve for \(y\):
Subtract \(5x\) from both sides: \(2y=-5x + 8\)
Divide each term by 2: \(y=-\frac{5}{2}x + 4\)
The slope \(m =-\frac{5}{2}\) (negative slope, so the line should be decreasing from left to right) and the y - intercept \(b = 4\)? Wait, no, wait. Wait, when we divide \(8\) by \(2\), we get \(y=-\frac{5}{2}x+4\)? Wait, no, \(8\div2 = 4\)? Wait, no, \(5x+2y = 8\), \(2y=-5x + 8\), \(y=-\frac{5}{2}x + 4\)? Wait, but let's check the y - intercept. When \(x = 0\), \(y=4\)? Wait, no, in the equation \(5x+2y = 8\), when \(x = 0\), \(2y=8\), \(y = 4\). When \(y=0\), \(5x=8\), \(x=\frac{8}{5}=1.6\)? Wait, no, the graphs in the options have x - intercepts at \(x = 4\) (for option A) and \(x=-4\) (for option B) and y - intercept at \(y = 2\) (for both? Wait, looking at the graphs:

In option A: The line crosses the y - axis at \(y = 2\) and x - axis at \(x = 4\). Let's check if the equation \(5x+2y = 8\) passes through \((0,2)\) and \((4,0)\).

For \((0,2)\): Left - hand side (LHS): \(5(0)+2(2)=4
eq8\). Wait, that's not right. Wait, maybe I made a mistake in rewriting. Wait, \(5x+2y = 8\), let's solve for \(y\) again:

\(2y=-5x + 8\)

\(y=-\frac{5}{2}x+4\). So when \(x = 0\), \(y = 4\); when \(y = 0\), \(5x=8\), \(x=\frac{8}{5}=1.6\). But the graphs in the options have y - intercept at \(y = 2\) and x - intercept at \(x = 4\) (option A) or \(x=-4\) (option B). Wait, maybe there is a typo in my calculation. Wait, maybe the original equation is \(x+2y = 8\)? No, the problem says \(5x + 2y=8\). Wait, maybe the graphs are mislabeled, or maybe I misread the equation. Wait, looking at the graphs:

Option A: The line goes through \((0,2)\) and \((4,0)\). Let's check if \(5x+2y = 8\) passes through \((4,0)\): \(5(4)+2(0)=20
eq8\). Not good. Wait, maybe the equation is \(x + 2y=8\)? Then \(2y=-x + 8\), \(y=-\frac{1}{2}x + 4\). No, still not. Wait, maybe the equation is \(x+2y = 4\)? Then \(2y=-x + 4\), \(y=-\frac{1}{2}x + 2\). Then when \(x = 0\), \(y = 2\); when \(y = 0\), \(x = 4\). Ah! That matches the graph in option A. Maybe there is a typo in the problem, and the equation is \(x + 2y=8\)? No, the problem says \(5x+2y = 8\). Wait, maybe the user made a typo, but looking at the graphs:

Option A: The line has a negative slope (decreasing from left to right), crosses y - axis at \(y = 2\) and x - axis at \(x = 4\). Let's check the equation of the line in option A. The slope \(m=\frac{0 - 2}{4-0}=\frac{-2}{4}=-\frac{1}{2}\). So the equation is \(y=-\frac{1}{2}x + 2\). Multiply both sides by 2: \(2y=-x + 4\), or \(x+2y = 4\). But the problem says \(5x+2y = 8\). Wait, maybe the problem is \(x + 2y=4\), but regardless, among the two options, option A has a negative slope (decreasing line) and option B has a positive slope (increasing line). The equation \(5x+2y = 8\) has a negative slope (\(y=-\frac{5}{2}x + 4\)), so the line should be decreasing. So option A is the correct one because it has a negative slope (the line goes down from left to right), while option B has a positive slope (goes up from left to right).

Answer:

A. The line with negative slope (decreasing from left to right), crossing y - axis at \(y = 2\) and x - axis at \(x = 4\)