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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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    Follow Andre's answer you get $s=9, t=\frac{25}{4}$2012-05-06

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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.