{x+2y=-1, -x+3y= -2

Question

a)[-7 3]
b)[3 -7]
c)[7 -3]
d)no solution

Answer 1

To solve this system of equations, we can use the method of elimination.

First, let's multiply the first equation by 3 and the second equation by 2 to get:
3(x + 2y) = 3(-1) -> 3x + 6y = -3
2(-x + 3y) = 2(-2) -> -2x + 6y = -4

Now we can add the two equations together:
(3x + 6y) + (-2x + 6y) = -3 + (-4)
x + 12y = -7

Since the coefficients of x are the same in both equations, they cancel out when we add them together.

Now, multiplying the second equation by 3 gives us:
3(-x + 3y) = 3(-2) -> -3x + 9y = -6

Adding this equation to x + 12y = -7, we get:
(-3x + 9y) + (x + 12y) = -6 + (-7)
-3x + x + 9y + 12y = -6 - 7
-2x + 21y = -13

Now, let's solve this equation for x:
-2x + 21y = -13 -> -2x = -13 - 21y -> x = (13 + 21y)/2

We now have expressions for both x and y in terms of y. We can substitute the expression for x into the first equation:

x + 2y = -1
(13 + 21y)/2 + 2y = -1
13 + 21y + 4y = -2
25y = -15
y = -15/25
y = -3/5

Substituting this value of y back into the expression for x:

x = (13 + 21(-3/5))/2
x = (13 - 63/5)/2
x = (65/5 - 63/5)/2
x = 2/5

Therefore, the solution to the system of equations is x = 2/5 and y = -3/5.

The answer is d) no solution.