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Multi-step equation with radical
Solve √(2-6y) - √(3) = 0.
3 Answers By Expert Tutors

Lale A. answered 08/10/24
Algebra Expertise with a Strategic, Student-Centered Approach
To solve the equation \sqrt{2 - 6y} - \sqrt{3} = 0, follow these steps:
1. Isolate the square root term:
\sqrt{2 - 6y} = \sqrt{3}
2. Square both sides to eliminate the square root:
\sqrt{2 - 6y})^2 = (\sqrt{3})^2
2 - 6y = 3
3. Solve for y:
Subtract 2 from both sides:
-6y = 3 - 2
-6y = 1
Divide by -6:
y = \frac{1}{-6}
y = -\frac{1}{6}
So, the solution is y = -\frac{1}{6} .

Michael M. answered 08/10/24
Caltech grad teaches math and coding the way your brain wants to learn
As your tutor, I'll teach you both some facts that you need to have memorized, and some guiding rules of thumb or guiding intuitions that you can use to understand this problem and to solve other problems, even entirely new ones.
To solve this equation, we need to isolate 'y' on one side of the equation by itself. One way to develop a guiding intuition for how to isolate 'y' is to identify how it is not currently isolated. Here are the ways it is not isolated:
- 'y' is inside a radical sign
- There is another term inside that radical sign as well.
- There are other terms on the same side as 'y'
Let's tackle these problems one at a time.
For the radical sign, we need a fact, which is that you should square each side of the equation, possibly manipulating it first.
For the moment let's square the equation without manipulating it.
(2-6y) - 2√(2-6y)√3 + 3 = 0
Take a look at that. Here's another guiding intuition: when the equation is getting more complicated as you manipulate it, you may be on the wrong track. Here the equation is definitely more complicated and we haven't gotten rid of the radical signs!
Let's try manipulating the equation first:
√(2-6y) = √3
Now square:
2-6y = 3
Much simpler! No radicals! We're on the right track. Now we solve this equation using Algebra 1, and we get
y = -1/6
Here's another fact: when you solve an equation by squaring both sides, you may create extraneous solutions, which means that one or more of the solutions you found may not work if you plug them back into the original equation. Here, -1/6 checks out, so this is the answer.
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