1)
If (2x+3y)(3y+2x)=z, according to the given information, and x=3.2 and z=457.96, just plug those values in to solve for y:
(2x+3y)(3y+2x)=z | Plug in the appropriate values for the given variables of x and z. | ||
(2∗3.2+3y)(2∗3.2+3y)=457.96 | Simplify what is inside the parentheses first. | ||
(6.4+3y)(6.4+3y)=457.96 | You might notice that both the multiplicand and multiplier are the same, which means that we can make this equation a tad simpler. | ||
(6.4+3y)2=457.96 | Take the square root of both sides. Of course, this breaks the equation up into its positive and negative answer. | ||
6.4+3y=±√457.96 | Although it may not be obvious, the square root of happens to work out nicely. | ||
6.4+3y=±21.4 | To solve for y, we must break up the equation. | ||
| Now, subtract by 6.4 in both equations. | ||
| Divide by 3 on both sides. | ||
| Both of these y-values satisfy the equation, and these are the solutions. | ||
2)
This is a system of equations. I usually refrain from using the elimination method here because it is difficult to showcase. Therefore, I will use the substitution method.
I will solve for y in equation 2:
−2x−3y=2 | Add 2x to both sides. |
−3y=2x+2 | Divide by -3 to isolate y. |
y=−2x+23 | |
Plug this value for y into equation 1 and then solve for x.
3x+5y=−2 | Plug in the value for y that was determined from the previous equation. |
3x+5∗2x+2−3=−2 | Do the multiplication first to simplify this monstrosity. |
5∗2x+2−3=5(2x+2)−3=10x+10−3 | Now, reinsert this back into the original equation. |
3x+10x+10−3=−2 | Multiply by -3 on all sides to get rid of the fraction. |
−9x+10x+10=6 | Combine the like terms on the left hand side. |
x+10=6 | Subtract 10 on both sides. |
x=−4 | |
Now, plug x=-4 into either equation and solve for y. I'll choose equation 2 because it look easier to do:
−2x−3y=2 | Substitute all x's for -4. |
−2∗−4−3y=2 | |
8−3y=2 | Subtract by 8 on both sides. |
−3y=−6 | Divide by -3 to isolate y. |
y=2 | |
Therefore, the coordinate where both lines intersect is (−4,2).