Question

equations in standard form.
$5x + y = 25$,
$x + 5y = -25$
keisha used the graphing calculator and identified the solution as $(6, -6)$. is she correct? if not, what was her mistake?
○ yes, keisha is correct.
○ no. she switched the x and y values
○ no. she only estimated instead of hovering over the intersection to find the exact point.
○ no. she picked a point on one line instead of the intersection point.

Explanation:

Step1: Check (6, -6) in first equation

Substitute \(x = 6\), \(y=-6\) into \(5x + y = 25\).
Left - hand side (LHS) \(=5\times6+(-6)=30 - 6 = 24
eq25\) (RHS).

Step2: Check (6, -6) in second equation

Substitute \(x = 6\), \(y = - 6\) into \(x+5y=-25\).
LHS \(=6+5\times(-6)=6 - 30=-24
eq - 25\) (RHS).

Step3: Analyze the mistake

A solution to a system of equations must satisfy both equations. Since \((6,-6)\) does not satisfy either equation, Keisha made a mistake in identifying the intersection point. The option "No. She only estimated instead of hovering over the intersection to find the exact point" is the most appropriate because if she had found the exact intersection (by hovering or solving algebraically), she would have seen that \((6,-6)\) is not the solution. (We can also solve the system algebraically: Multiply the first equation by 5: \(25x + 5y=125\), subtract the second equation \(x + 5y=-25\) from it: \(24x=150\), \(x=\frac{150}{24}=\frac{25}{4} = 6.25\), then substitute \(x=\frac{25}{4}\) into \(x + 5y=-25\), \(\frac{25}{4}+5y=-25\), \(5y=-25-\frac{25}{4}=-\frac{125}{4}\), \(y =-\frac{25}{4}=-6.25\). So the exact solution is \((\frac{25}{4},-\frac{25}{4})\) or \((6.25,-6.25)\), which is close to \((6,-6)\) but not exact, indicating an estimation error.)

Answer:

No. She only estimated instead of hovering over the intersection to find the exact point.