Please help again!
Find the ordered pair that satisfies the system of equations
\(\begin{align*} x+3y &= -4,\\ 3x -7y &= 36. \end{align*}\)
subtract 3x the first equation from the second equation
\((3x-7y)-3(x+3y) = 36 - 3(-4)\\ -16y=48\\ y=-3\\ \text{use this value of y to obtain x}\\ x + 3(-3) = -4 \\ x - 9 = -4\\ x = 5\\ (x,y) = (5,-3) \)
Substitution: x+3y=-4, so x=-4-3y and 3(-4-3y)-7y=36, -12-9y-7y=36, -9y-7y=48, -16y=48, y=-3. and x=x+3(-3)=-4, x-9=-4, x=5. Thus, the ordered pair is (5, -3).
Elimination: We can also show this by elimination: multiply the first equation by 3, which yields, 3x+9y=-12 and 3x-7y=36. Subtract, 16y=-48, y=-3. Plugging the value of y back in to find the value of x, we get x+3(-3)=-4, x-9=-4, x=5. Thus, the ordered pair is (5, -3).