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tayaiv

  • one year ago

Simultaneous equations: 6 = b√ 2 3a = b How can I find a?

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  1. zpupster
    • one year ago
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    \[6=b \sqrt{2}\] divide both sides by sqrt 2 then you have b plug it in the 3a = b then divide both sides by 3 now you have a

  2. tayaiv
    • one year ago
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    What happens to 6 in the first equation when both sides are divided by sqrt 2? Why is only b left?

  3. anonymous
    • one year ago
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    What you have to get is that algebra is like a set of scales. Whatever you do to one side of an equation you do to the other, and the only way to solve for a is to find b. We know 6 = b *sqrt(2). We want b by itself, to do that we get rid of the sqrt(2). By dividing both sides of the equation by sqrt(2) it leaves b= 6/(sqrt(2))

  4. zpupster
    • one year ago
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    b= 6/(sqrt(2)) so plug it in the 2nd equation 3a = 6/(sqrt(2)) divide both sides by 3, now you have a

  5. tayaiv
    • one year ago
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    Oh, got it! Thank you :)

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