rationalizing the numerator, or namely, "getting rid of that pesky radical at the top".
we simply multiply top and bottom by a value that will take out the radicand in the numerator.
![\bf \cfrac{\sqrt[3]{144x}}{\sqrt[3]{y}}~~ \begin{cases} 144=2\cdot 2\cdot 2\cdot 2\cdot 3\cdot 3\\ \qquad 2^3\cdot 18 \end{cases}\implies \cfrac{\sqrt[3]{2^3\cdot 18x}}{\sqrt[3]{y}}\implies \cfrac{2\sqrt[3]{ 18x}}{\sqrt[3]{y}} \\\\\\ \cfrac{2\sqrt[3]{ 18x}}{\sqrt[3]{y}}\cdot \cfrac{\sqrt[3]{(18x)^2}}{\sqrt[3]{(18x)^2}}\implies \cfrac{2\sqrt[3]{(18x)(18x)^2}}{\sqrt[3]{(y)(18x)^2}}\implies \cfrac{2\sqrt[3]{(18x)^3}}{\sqrt[3]{18^2x^2y}}](https://tex.z-dn.net/?f=%5Cbf%20%5Ccfrac%7B%5Csqrt%5B3%5D%7B144x%7D%7D%7B%5Csqrt%5B3%5D%7By%7D%7D~~%0A%5Cbegin%7Bcases%7D%0A144%3D2%5Ccdot%202%5Ccdot%202%5Ccdot%202%5Ccdot%203%5Ccdot%203%5C%5C%0A%5Cqquad%202%5E3%5Ccdot%2018%0A%5Cend%7Bcases%7D%5Cimplies%20%5Ccfrac%7B%5Csqrt%5B3%5D%7B2%5E3%5Ccdot%20%2018x%7D%7D%7B%5Csqrt%5B3%5D%7By%7D%7D%5Cimplies%20%5Ccfrac%7B2%5Csqrt%5B3%5D%7B%20%2018x%7D%7D%7B%5Csqrt%5B3%5D%7By%7D%7D%0A%5C%5C%5C%5C%5C%5C%0A%5Ccfrac%7B2%5Csqrt%5B3%5D%7B%20%2018x%7D%7D%7B%5Csqrt%5B3%5D%7By%7D%7D%5Ccdot%20%5Ccfrac%7B%5Csqrt%5B3%5D%7B%2818x%29%5E2%7D%7D%7B%5Csqrt%5B3%5D%7B%2818x%29%5E2%7D%7D%5Cimplies%20%5Ccfrac%7B2%5Csqrt%5B3%5D%7B%2818x%29%2818x%29%5E2%7D%7D%7B%5Csqrt%5B3%5D%7B%28y%29%2818x%29%5E2%7D%7D%5Cimplies%20%5Ccfrac%7B2%5Csqrt%5B3%5D%7B%2818x%29%5E3%7D%7D%7B%5Csqrt%5B3%5D%7B18%5E2x%5E2y%7D%7D)
![\bf \cfrac{2(18x)}{\sqrt[3]{324x^2y}}~~ \begin{cases} 324=2\cdot 2\cdot 3\cdot 3\cdot 3\cdot 3\\ \qquad 12\cdot 3^3 \end{cases}\implies \cfrac{36x}{\sqrt[3]{12\cdot 3^3x^2y}} \\\\\\ \cfrac{36x}{3\sqrt[3]{12x^2y}}\implies \cfrac{12x}{\sqrt[3]{12x^2y}}](https://tex.z-dn.net/?f=%5Cbf%20%5Ccfrac%7B2%2818x%29%7D%7B%5Csqrt%5B3%5D%7B324x%5E2y%7D%7D~~%0A%5Cbegin%7Bcases%7D%0A324%3D2%5Ccdot%202%5Ccdot%203%5Ccdot%203%5Ccdot%203%5Ccdot%203%5C%5C%0A%5Cqquad%2012%5Ccdot%203%5E3%0A%5Cend%7Bcases%7D%5Cimplies%20%5Ccfrac%7B36x%7D%7B%5Csqrt%5B3%5D%7B12%5Ccdot%203%5E3x%5E2y%7D%7D%0A%5C%5C%5C%5C%5C%5C%0A%5Ccfrac%7B36x%7D%7B3%5Csqrt%5B3%5D%7B12x%5E2y%7D%7D%5Cimplies%20%5Ccfrac%7B12x%7D%7B%5Csqrt%5B3%5D%7B12x%5E2y%7D%7D)
1. V = lwh
V = (8)(4)(5)
V = (32)(5)
V = 160 units³
2. V = lwh + lwh
V = (4)(2)(1) + (3)(1)(1)
V = (8)(1) + (3)(1)
V = 8 + 3
V = 11 units³
3. V = leh + s³
V = (9)(3)(4) + (3)³
V = (27)(4) + 27
V = 108 + 27
V = 135 units³
4. V = lwh
V = (3)(7)(2)
V = (21)(2)
V = 42 units³
Answer:
No solutions
Step-by-step explanation:
In this question, you would be solving for x.
Solve:
5 + 2(3 + 2x) = x + 3(x + 1)
Use the distributive property.
5 + 6 + 4x = x + 3x + 3
Combine like terms.
11 + 4x = 4x + 3
Subtract 4x from both sides.
11 = 3
Since we don't have an "x value", there are no solutions.
"No solutions" would be your answer.
So I think you should set up an equation like this: 14.95x+16.88y=131.18 , x+y=8
so x=8-y, plug this into the original equation:
14.95(8-y)+16.88y=131.18
Evaluate: 119.60-14.95y+16.88y=131.18
Combine like terms: 119.60+1.93y=131.18
Subtract 119.60 from both sides. You get: 1.93y=11.58
Divide both sides by 1.93. You get y=6
Plug into second equation: x+y=8 ---> x+6=8 and solve for x. x=2
Christine bought 2 video games and 6 CDs. 14.95(2)+16.88(6)=131.18
