(2x + 3y = 12) x (-2)
(4x - 3y = 6) x 1
-4x - 6y = -24
4x - 3y = 6
You can cancel out the x values by adding the two equations together.
(-4x + 4x) + (-6y - 3y) = (-24 + 6)
-9y = -18
y = 2
Solve for x now...
4x - 3(2) = 6
4x - 6 = 6
4x = 12
x = 3
Check... (x = 3, y = 2)
2(3) + 3(2) = 12
6 + 6 = 12
12 = 12 <- this works!
4(3) - 3(2) = 6
12 - 6 = 6
6 = 6 <- this works!
Answer:
Step-by-step explanation:
1). Step 4:
[Since,
]
![x=\sqrt[3]{5\times 5\times 5\times 5}](https://tex.z-dn.net/?f=x%3D%5Csqrt%5B3%5D%7B5%5Ctimes%205%5Ctimes%205%5Ctimes%205%7D)
Step 5:
![x=\sqrt[3]{(5)^3\times 5}](https://tex.z-dn.net/?f=x%3D%5Csqrt%5B3%5D%7B%285%29%5E3%5Ctimes%205%7D)
![x=\sqrt[3]{5^3}\times \sqrt[3]{5}](https://tex.z-dn.net/?f=x%3D%5Csqrt%5B3%5D%7B5%5E3%7D%5Ctimes%20%5Csqrt%5B3%5D%7B5%7D)
2). He simplified the expression by removing exponents from the given expression.
3). Let the radical equation is,

Step 1:

Step 2:

Step 3:

Step 4:

4). By substituting
in the original equation.



There is no extraneous solution.
Answer:
mutiply
Step-by-step explanation:
I think the answer is B. but I'm not sure... Sorry if this doesn't help!