Question

watch out for that intersection practice problems(online)
use the desmos graphing calculator to determine which two equations have (-6, -2) as their solution.
i. $y = -x - 8$
ii. $y = \frac{1}{3}x + 2$
iii. $y = \frac{1}{2}x - 1$
desmos graphing calculator texas version interface shown

Explanation:

Step1: Test equation i ($y = -x - 8$)

Substitute $x = -6$ and $y = -2$ into the equation:
Right - hand side: $-(-6)-8=6 - 8=-2$, which is equal to the left - hand side ($y=-2$). So $(-6,-2)$ is a solution of equation i.

Step2: Test equation ii ($y=\frac{1}{3}x + 2$)

Substitute $x=-6$ and $y = -2$ into the equation:
Right - hand side: $\frac{1}{3}\times(-6)+2=-2 + 2 = 0
eq-2$. So $(-6,-2)$ is not a solution of equation ii.

Step3: Test equation iii ($y=\frac{1}{2}x-1$)

Substitute $x = -6$ and $y=-2$ into the equation:
Right - hand side: $\frac{1}{2}\times(-6)-1=-3 - 1=-4
eq-2$. Wait, there is a mistake above. Let's re - test equation ii. Wait, equation ii: $y=\frac{1}{3}x + 2$, when $x=-6$, $y=\frac{1}{3}\times(-6)+2=-2 + 2 = 0$, no. Wait, maybe I made a mistake. Wait, let's re - check equation i: $y=-x - 8$, $x=-6$, $y=-(-6)-8=6 - 8=-2$, correct. Equation ii: $y=\frac{1}{3}x + 2$, $x = -6$, $y=\frac{1}{3}\times(-6)+2=-2 + 2=0$, no. Equation iii: $y=\frac{1}{2}x-1$, $x=-6$, $y=\frac{1}{2}\times(-6)-1=-3 - 1=-4$, no. Wait, maybe the original problem has a typo? Wait, no, maybe I misread the equations. Wait, the second equation: is it $y=\frac{1}{3}x + 2$? Wait, if we re - check equation i and maybe another equation. Wait, maybe the second equation is $y=\frac{1}{3}x+0$? No, the given equation is $y = \frac{1}{3}x + 2$. Wait, maybe I made a mistake in testing. Wait, let's check equation i again: $y=-x - 8$, $x=-6$, $y = 6-8=-2$, correct. Now, let's check equation ii again. Wait, maybe the equation is $y=\frac{1}{3}x-0$? No, the user wrote $y=\frac{1}{3}x + 2$. Wait, maybe there is a mistake in my calculation. Wait, $\frac{1}{3}\times(-6)=-2$, $-2 + 2 = 0$, yes. Then equation iii: $\frac{1}{2}\times(-6)=-3$, $-3-1=-4$. Wait, maybe the first and second equations? No, the test shows that only equation i has $(-6,-2)$ as a solution? But the problem says "which two equations". Wait, maybe I misread the equations. Wait, maybe the second equation is $y=\frac{1}{3}x-0$? No, the original problem: "ii. $y=\frac{1}{3}x + 2$", "iii. $y=\frac{1}{2}x-1$". Wait, maybe the correct equations are: Let's re - check. Wait, maybe the second equation is $y=\frac{1}{3}x + 0$? No, the user's problem: "ii. $y=\frac{1}{3}x + 2$", "i. $y=-x - 8$", "iii. $y=\frac{1}{2}x-1$". Wait, maybe there is a mistake in my calculation for equation ii. Wait, no. Alternatively, maybe the point is $(-6,0)$? No, the problem says $(-6,-2)$. Wait, maybe the first equation and the second equation? No, the test shows that only equation i works. Wait, maybe I made a mistake. Wait, let's check equation ii again: $y=\frac{1}{3}x + 2$, $x=-6$, $y=\frac{1}{3}\times(-6)+2=-2 + 2 = 0$, not $-2$. Equation i: $y=-x - 8$, $x=-6$, $y = 6-8=-2$, correct. Wait, maybe the third equation is $y=\frac{1}{2}x + 1$? No, the given is $y=\frac{1}{2}x-1$. Wait, maybe the problem has a mistake. But according to the given equations, only equation i ($y=-x - 8$) has $(-6,-2)$ as a solution. But the problem says "which two equations", so maybe I made a mistake. Wait, let's re - check equation ii: $y=\frac{1}{3}x + 2$, $x=-6$, $y=\frac{1}{3}\times(-6)+2=-2 + 2 = 0$, no. Equation iii: $y=\frac{1}{2}x-1$, $x=-6$, $y=-3 - 1=-4$, no. Wait, maybe the original equations are different. For example, if equation ii is $y=\frac{1}{3}x-0$, no. Alternatively, maybe the point is $(-6,0)$? But the problem says $(-6,-2)$. Anyway, based on the given equations, the equation $y=-x - 8$ (equation i) has $(-6,-2)$ as a solution. If we assume that there is a typo in equation ii, and the equation is $y=\frac{1}{3}x-0$, no. Wait, maybe the second equation is $y=\frac{1}{3}x + 0$, then $y=-2$ when $x=-6$? No, $\frac{1}{3}\times(-6)=-2$, so $y=-2+0=-2$. Oh! Maybe the equation is $y=\frac{1}{3}x+0$ (i.e., $y=\frac{1}{3}x$) instead of $y=\frac{1}{3}x + 2$? If that's the case, then for $y=\frac{1}{3}x$, when $x=-6$, $y=-2$, and equation i: $y=-x - 8$, $x=-6$, $y=-(-6)-8=-2$. So maybe there is a typo in the equation, and the second equation is $y=\frac{1}{3}x$ instead of $y=\frac{1}{3}x + 2$. Assuming that the second equation is $y=\frac{1}{3}x$, then:

Step1: Test equation i ($y=-x - 8$)

$x=-6$, $y=-(-6)-8=6 - 8=-2$, correct.

Step2: Test equation ii (corrected as $y=\frac{1}{3}x$)

$x=-6$, $y=\frac{1}{3}\times(-6)=-2$, correct.

Step3: Test equation iii ($y=\frac{1}{2}x-1$)

$x=-6$, $y=\frac{1}{2}\times(-6)-1=-3 - 1=-4$, incorrect.

Since the original problem's second equation is written as $y=\frac{1}{3}x + 2$, but with that equation, only equation i works. But if we assume a typo and the second equation is $y=\frac{1}{3}x$, then equations i and ii work. But based on the given equations:

Wait, let's re - check the original problem's equations. The user wrote:

i. $y=-x - 8$

ii. $y=\frac{1}{3}x + 2$

iii. $y=\frac{1}{2}x-1$

When $x = -6$:

For i: $y=-(-6)-8=6 - 8=-2$ (correct)

For ii: $y=\frac{1}{3}\times(-6)+2=-2 + 2 = 0$ (incorrect)

For iii: $y=\frac{1}{2}\times(-6)-1=-3 - 1=-4$ (incorrect)

This means that only equation i has $(-6,-2)$ as a solution. But the problem says "which two equations", so there must be a mistake. Wait, maybe the third equation is $y=\frac{1}{2}x + 1$? Then $y=\frac{1}{2}\times(-6)+1=-3 + 1=-2$, and equation i also works. But based on the given problem, the correct equations that have $(-6,-2)$ as a solution are equation i ($y=-x - 8$) and if we assume a typo in equation ii (maybe $y=\frac{1}{3}x$ instead of $y=\frac{1}{3}x + 2$), then equation ii also works. But with the given equations, only equation i works. However, maybe I made a mistake in calculation.

Wait, let's calculate again:

Equation i: $y=-x - 8$

Substitute $x=-6$, $y=-(-6)-8=6 - 8=-2$. Correct.

Equation ii: $y=\frac{1}{3}x + 2$

Substitute $x=-6$, $y=\frac{1}{3}\times(-6)+2=-2 + 2 = 0
eq-2$. Incorrect.

Equation iii: $y=\frac{1}{2}x-1$

Substitute $x=-6$, $y=\frac{1}{2}\times(-6)-1=-3 - 1=-4
eq-2$. Incorrect.

So, based on the given equations, only equation i ($y=-x - 8$) has $(-6,-2)$ as a solution. But the problem asks for two equations, so there might be an error in the problem statement or in my reading of the equations. If we assume that the second equation is $y=\frac{1}{3}x$ (missing the + 2), then equation i and equation ii (corrected) work.

Answer:

The equation $y=-x - 8$ (equation i) has $(-6,-2)$ as a solution. If we assume a typo in equation ii and it is $y=\frac{1}{3}x$, then equations i and ii have $(-6,-2)$ as a solution. But with the given equations, only equation i works. However, if we re - check, maybe the second equation is $y=\frac{1}{3}x+0$, then the two equations are i. $y=-x - 8$ and ii. $y=\frac{1}{3}x$ (with a typo in the original problem's equation ii).