To solve for x, you must use the square root and remember to split the answers into 2:
\((x+3)^2=7\) | Take the square root of both sides |
\(x+3=\sqrt{7}\) | However, remember that the absolute value equation splits your answer into the negative and positive answer |
\(x+3=\pm\sqrt{7}\) | Subtract 3 on both sides |
\(x=\pm\sqrt{7}-3\) | |
\(x\approx-0.35425\hspace{1mm}\text{and}\hspace{1mm}x\approx-5.64575\) | Here are your answers rounded, but you want to keep your answer as it is above because that is the exact value. |
To solve for x and y, first solve for a variable and plug it into the other equation. Since you specified solving by substitution, I'll use that method. Here are your 2 equations:
1. \(2x+4y=-32\)
2. \(-3x+y=6\)
I'll solve for y in the second equation because it has a coefficient of 1.
\(-3x+y=6\) | To isolate y, add 3x to both sides. |
\(y=6+3x\) | |
Now that I have solved for y in one equation, substitute y in the other.
\(2x+4y=-32\) | In the previous calculation, we deduced that y=6+3x, so replace y. |
\(2x+4(6+3x)=-32\) | Distribute the 4 into the parentheses to ease the simplification process. |
\(2x+24+12x=-32\) | Combine the like terms, specifically 2x and 12x. |
\(14x+24=-32\) | Subtract 24 on both sides. |
\(14x=-56\) | Divide by 14 on both sides to isolate x. |
\(x=-4\) | |
Plug x into an equation and solve for y. I'll plug it into equation 2
\(-3x+y=6\) | Substitute the calculated value for x, -4. |
\(-3(-4)+y=6\) | Simplify -3*-4. |
\(12+y=6\) | Subtract 12 on both sides |
\(y=-6\) | |
Therefore, the solution set is (-4,-6)