# Find $s$ and $t$ with substitution?

First equation is $$5\sqrt{s} - 2\sqrt{t} = 10.$$

Second equation is $$2s = 4t - 7.$$

It should be done with substituition. Please write down your way for solution. Thanks.

I tried and stick to this point.

squared the first one. $$25s = 4t + 100 + 40\sqrt{t}.$$ Then 2nd eq (4t = 2s +7) in first.. that got me here $$23s = 107 + 40\sqrt{t}.$$

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What have you tried? –  Amitabh Udayiman May 6 '12 at 18:21
$5\sqrt s-\sqrt{4t}=10\rightarrow5\sqrt s=10+\sqrt{2s+7}$. Square both sides up: $23s-107=20\sqrt{2s+7}$; then square again. –  wangdw May 6 '12 at 18:23
I did as you said. but I got 529s^2 - 5722s + 8649 = 0.. whereby 529 and 8649 are square. but 5722 is not 23*93*2 to make it binom. what should I do? –  Casper May 6 '12 at 18:38
Follow Andre's answer you get $s=9, t=\frac{25}{4}$ –  Kirthi Raman May 6 '12 at 20:35

We have $2\sqrt{t}=5\sqrt{s}-10$. Square both sides.

We get $4t=(5\sqrt{s}-10)^2$. But $4t=2s+7$.

So we are at $2s+7=(5\sqrt{s}-10)^2$.

Let $w=\sqrt{s}$. Then $2w^2+7=5^2(w-2)^2=25w^2-100w+100$. Now we can use the Quadratic Formula, though in fact the polynomial $23w^2-100w+93$ factors pleasantly, and we get $w=3$ or $w=31/23$, giving $s=9$ or $s=(31/23)^2$. Now we can find the accompanying $t$.

Remember that we squared during the solving process. This may have resulted in "extraneous" roots that are not solutions of the original system. So do check.

Remark: Alternately, we could write $5\sqrt{s}=2\sqrt{t}+10$ and square both sides. The only reason I did not do that is that $s=(4t-7)/2$, so substitution produces fractions. However, multiplying through by $4$ clears the fractions. In principle the details are pretty similar to the details for the approach I chose, with somewhat more unpleasant numbers.

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