To solve a multi-variable problem, you need to isolate single variables. To do this, there are two ways to do it. One way you eliminate one of the variables and the other you substitute.
In this example, you are being asked to eliminate one of the variables. To do this, you multiply 2x+2y= 12 by 2 and get 4x+4y = 24.
See how now both equations contain a 4y?
Using this you can subtract one equation from the other.
5x+4y=28
- 4x+4y=24
And then you get x=4
To get the other variable, you substitute 4 in for x (in one of the equations) and then solve for y.
2(4) + 2y =12
8+2y= 12
2y=4
y=2
Therefore, x = 4 and y = 2
we know the triangle is an isosceles, so it has twin sides, and then the hypotenuse, since it's a right-triangle, we can just use the pythagorean theorem for that.
![\bf \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies c=\sqrt{a^2+b^2} \qquad \begin{cases} c=hypotenuse\\ a=\stackrel{adjacent}{3\sqrt{2}}\\ b=\stackrel{opposite}{3\sqrt{2}}\\ \end{cases} \\\\\\ c=\sqrt{(3\sqrt{2})^2+(3\sqrt{2})^2}\implies c=\sqrt{(3^2\cdot 2)+(3^2\cdot 2)}\implies c=\sqrt{18+18} \\\\\\ c=\sqrt{36}\implies c=6](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Busing%20the%20pythagorean%20theorem%7D%20%5C%5C%5C%5C%20c%5E2%3Da%5E2%2Bb%5E2%5Cimplies%20c%3D%5Csqrt%7Ba%5E2%2Bb%5E2%7D%20%5Cqquad%20%5Cbegin%7Bcases%7D%20c%3Dhypotenuse%5C%5C%20a%3D%5Cstackrel%7Badjacent%7D%7B3%5Csqrt%7B2%7D%7D%5C%5C%20b%3D%5Cstackrel%7Bopposite%7D%7B3%5Csqrt%7B2%7D%7D%5C%5C%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5C%5C%20c%3D%5Csqrt%7B%283%5Csqrt%7B2%7D%29%5E2%2B%283%5Csqrt%7B2%7D%29%5E2%7D%5Cimplies%20c%3D%5Csqrt%7B%283%5E2%5Ccdot%202%29%2B%283%5E2%5Ccdot%202%29%7D%5Cimplies%20c%3D%5Csqrt%7B18%2B18%7D%20%5C%5C%5C%5C%5C%5C%20c%3D%5Csqrt%7B36%7D%5Cimplies%20c%3D6)
2,119,500
explanation: you multiply
Answer:
x = 6
Step-by-step explanation:
3x - 1 = 17
add 1 to both sides:
3x = 18
divide by 3:
x = 6
I think it is 16 because I just divided it, I’m sorry if you will get it wrong I’m not sure