Question

solve it by cross multiplication ​

Answer 1

Step-by-step explanation:

given: x/a=y/b

by cross multiplying...

bx=ay

so, x= ay/b, y= bx/a

hope it helps you!

Answer 2

Question:

$\sf{Solve \ for \ x \ and \ y : \ \dfrac{x}{a}=\dfrac{y}{b}}$

$\sf{ax+by=a^{2}+b^{2}}$

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Answer:

$\sf{The \ values \ of \ x \ and \ y \ are}$

$\sf{a \ and \ b \ respectively.}$

Given:

$\sf{The \ given \ equations \ are}$

$\sf{\dfrac{x}{a}=\dfrac{y}{b} \ and}$

$\sf{ax+by=a^{2}+b^{2}}$

To find:

$\sf{The \ value \ of \ x \ and \ y.}$

Solution:

$\sf{\dfrac{x}{a}=\dfrac{y}{b}...(1)}$

$\sf{ax+by=a^{2}+b^{2}...(2)}$

$\sf{From \ equation (1), \ we \ get}$

$\sf{\dfrac{x}{a}=\dfrac{y}{b}}$

$\sf{\leadsto{\dfrac{x}{y}=\dfrac{a}{b}}}$

$\sf{\leadsto{x=\dfrac{ay}{b}}}$

$\sf{Substitute \ x=\dfrac{ay}{b} \ in \ equation (2), \ we \ get}$

$\sf{a(\dfrac{ay}{b})+by=a^{2}+b^{2}}$

$\sf{\therefore{\dfrac{a^{2}y}{b}+by=a^{2}+b^{2}}}$

$\sf{\therefore{\dfrac{a^{2}y+b^{2}y}{b}=a^{2}+b^{2}}}$

$\sf{\therefore{\dfrac{y(a^{2}+b^{2})}{b}=a^{2}+b^{2}}}$

$\sf{\therefore{\dfrac{y}{b}=1}}$

$\boxed{\sf{\therefore{y=b}}}$

$\sf{Substitute \ y=b \ in \ equation (1), \ we \ get}$

$\sf{\therefore{\dfrac{x}{a}=\dfrac{b}{b}}}$

$\sf{\therefore{\dfrac{x}{a}=1}}$

$\boxed{\sf{\therefore{x=a}}}$

$\sf\purple{\tt{\therefore{The \ values \ of \ x \ and \ y \ are}}}$

$\sf\purple{\tt{a \ and \ b \ respectively.}}$

_____________________________

By Cross Multiplication:

$\sf{The \ given \ equations \ are}$

$\sf{\dfrac{x}{a}=\dfrac{y}{b}}$

$\sf{\therefore{bx=ay}}$

$\sf{\therefore{bx-ay+0=0...(1)}}$

$\sf{ax+by=a^{2}+b^{2}}$

$\sf{\therefore{ax+by-a^{2}-b^{2}=0...(2)}}$

$\sf{Here,}$

$\sf{a_{1}=b, \ a_{2}=a, \ b_{1}=-a, \ b_{2}=b,}$

$\sf{c_{1}=0, \ c_{2}=-a^{2}-b^{2}}$

$\sf{By \ formula}$

$\boxed{\sf{x=\dfrac{b_{1}c_{2}-b_{2}c_{1}}{a_{1}b_{2}-a_{2}b_{1}}, \ y=\dfrac{c_{1}a_{2}-c_{2}a_{1}}{a_{1}b_{2}-a_{2}b_{1}}}}$

$\sf{\leadsto{x=\dfrac{(-a)(-a^{2}-b^{2})-0}{b^{2}-(a)(-a)}}}$

$\sf{\therefore{x=\dfrac{a^{3}+ab^{2}}{a^{2}+b^{2}}}}$

$\sf{\therefore{x=\dfrac{a(a^{2}+b^{2})}{a^{2}+b^{2}}}}$

$\boxed{\sf{\therefore{x=a}}}$

$\sf{\leadsto{y=\dfrac{0-(-a^{2}-b^{2})(b)}{b^{2}+a^{2}}}}$

$\sf{\therefore{y=\dfrac{a^{2}+b^{2}(b)}{a^{2}+b^{2}}}}$

$\boxed{\sf{\therefore{y=b}}}$

$\sf\purple{\tt{\therefore{The \ value \ of \ x \ is \ a}}}$

$\sf\purple{\tt{and \ y \ is \ b.}}$