Is (−2, 6)%0D%0A(%0D%0A−%0D%0A2%0D%0A,%0D%0A %0D%0A6%0D%0A)%0D%0A a solution to the system of these linear equations: x+2y=10%0D%0A%0D%0A+%0D%0A2%0D%0A%0D%0A=%0D%0A10%0D%0A and 3x+y=0%0D%0A3%0D%0A%0D%0A+%0D%0A%0D%0A=%0D%0A0%0D%0A? Why?(1 point)%0D%0AResponses%0D%0A%0D%0ANo, because the graphs don’t intersect at (−2, 6)%0D%0A(%0D%0A−%0D%0A2%0D%0A,%0D%0A %0D%0A6%0D%0A)%0D%0A.%0D%0ANo, because the graphs don’t intersect at left parenthesis negative 2 comma 6 right parenthesis .%0D%0A%0D%0AYes, because the graphs intersect at (−2, 6)%0D%0A(%0D%0A−%0D%0A2%0D%0A,%0D%0A %0D%0A6%0D%0A)%0D%0A.%0D%0AYes, because the graphs intersect at left parenthesis negative 2 comma 6 right parenthesis .%0D%0A%0D%0AYes, because the graphs don’t intersect at (−2, 6)%0D%0A(%0D%0A−%0D%0A2%0D%0A,%0D%0A %0D%0A6%0D%0A)%0D%0A.%0D%0AYes, because the graphs don’t intersect at left parenthesis negative 2 comma 6 right parenthesis .%0D%0A%0D%0ANo, because the graphs intersect at (−2, 6)%0D%0A(%0D%0A−%0D%0A2%0D%0A,%0D%0A %0D%0A6%0D%0A)%0D%0A.%0D%0ANo, because the graphs intersect at left parenthesis negative 2 comma 6 right parenthesis .
No, because the graphs don't intersect at (-2, 6).
No, because the graphs don't intersect at (-2, 6).
To determine if the point (−2, 6) is a solution to the system of linear equations, we need to substitute the values of x and y into both equations and check if both equations are satisfied.
The given equations are:
1) x + 2y = 10
2) 3x + y = 0
Let's substitute the values x = −2 and y = 6 into both equations:
For equation 1:
(−2) + 2(6) = 10
-2 + 12 = 10
10 = 10 -- This equation is satisfied.
For equation 2:
3(−2) + 6 = 0
-6 + 6 = 0
0 = 0 -- This equation is also satisfied.
Therefore, when we substitute x = −2 and y = 6 into both equations, both equations are satisfied. Hence, the point (−2, 6) is a solution to the system of linear equations.
So, the correct response is:
Yes, because the graphs intersect at (−2, 6).