y = -5x + 24
y = 4x - 21
Since both of these equations are equal to Y, theyre equal to each other.
So we can make an equation with y = -5x + 24 in one side and y = 4x - 21 on the other.
-5x + 24 = 4x - 21
Now in order to get the value of x we need to isolate it in one side of the equation. We can do this by subtracting 24 from both sides of the equation:
-5x + 24 - 24 = 4x - 21 - 24
-5x = 4x - 45
Now we subtract 4x from both sides so the 4x shift to the other side
-5x - 4x = 4x - 4x - 45
-9x = -45
Finally divide both sides by -9 so x is by itself
(-9)÷(-9x) = -(45)÷(-9)
x = 5
Since we did all of this to BOTH sides of the equation, both sides are still equal to each other and the equation still is true.
Now apply x = 5 to either of the initial equations to find the value of Y
y = -5x + 24 or y = 4x - 21
(I'll do both but u only need one)
y = -5(5) + 24
y = -25 + 24
y = -1
y = 4(5) - 21
y = 20 - 21
y = -1
Either way, X is 5 and Y is -1
Answer (5, -1)
Answer:
your answer is 30 ..
hope this will help you!!
well, we know it's a rectangle, so that means the sides JK = IL and JI = KL, so
![\stackrel{JK}{3x+21}~~ = ~~\stackrel{IL}{6y}\implies 3(x+7)=6y\implies x+7=\cfrac{6y}{3} \\\\\\ x+7=2y\implies \boxed{x=2y-7} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{JI}{6y-6}~~ = ~~\stackrel{KL}{2x+20}\implies 6(y-1)=2(x+10)\implies \cfrac{6(y-1)}{2}=x+10 \\\\\\ 3(y-1)=x+10\implies 3y-3=x+10\implies \stackrel{\textit{substituting from the 1st equation}}{3y-3=(2y-7)+10} \\\\\\ 3y-3=2y+3\implies y-3=3\implies \blacksquare~~ y=6 ~~\blacksquare ~\hfill \blacksquare~~ \stackrel{2(6)~~ - ~~7}{x=5} ~~\blacksquare](https://tex.z-dn.net/?f=%5Cstackrel%7BJK%7D%7B3x%2B21%7D~~%20%3D%20~~%5Cstackrel%7BIL%7D%7B6y%7D%5Cimplies%203%28x%2B7%29%3D6y%5Cimplies%20x%2B7%3D%5Ccfrac%7B6y%7D%7B3%7D%20%5C%5C%5C%5C%5C%5C%20x%2B7%3D2y%5Cimplies%20%5Cboxed%7Bx%3D2y-7%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Cstackrel%7BJI%7D%7B6y-6%7D~~%20%3D%20~~%5Cstackrel%7BKL%7D%7B2x%2B20%7D%5Cimplies%206%28y-1%29%3D2%28x%2B10%29%5Cimplies%20%5Ccfrac%7B6%28y-1%29%7D%7B2%7D%3Dx%2B10%20%5C%5C%5C%5C%5C%5C%203%28y-1%29%3Dx%2B10%5Cimplies%203y-3%3Dx%2B10%5Cimplies%20%5Cstackrel%7B%5Ctextit%7Bsubstituting%20from%20the%201st%20equation%7D%7D%7B3y-3%3D%282y-7%29%2B10%7D%20%5C%5C%5C%5C%5C%5C%203y-3%3D2y%2B3%5Cimplies%20y-3%3D3%5Cimplies%20%5Cblacksquare~~%20y%3D6%20~~%5Cblacksquare%20~%5Chfill%20%5Cblacksquare~~%20%5Cstackrel%7B2%286%29~~%20-%20~~7%7D%7Bx%3D5%7D%20~~%5Cblacksquare)
(2+4)/(-2-d) = -2
6 = -2(-2-d)
6 = 4 + 2d
2 = 2d
1 = d
