If $$(2x - y) : (5x + 3y) = 3 : 8$$, then $$(x^2 + y^2) : (x^2 - y^2) =$$
$$(2x - y) : (5x + 3y) = 3 : 8$$
$$\frac{(2x-y)}{(5x+3y)}\ =\ \frac{3}{8}$$
16x-8y = 15x+9y
16x-15x =Â 9y+8y
x = 17y
$$\frac{(x^2+y^2)}{(x^2-y^2)}$$ = $$\frac{\left(\left(17y\right)^2+y^2)\right)}{\left(\left(17y\right)^2-y^2)\right)}$$Â [putting the value of 'x'.]
=Â $$\frac{289y^2+y^2}{289y^2-y^2}$$
= $$\frac{290y^2}{288y^2}$$
=Â $$\frac{145}{144}$$
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