How to solve each of the following pairs of simultaneous equations for x and y? 1) $a x + y = c and x + b y = d$ $ax + y = c and x+by=d$

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Answer 1 · 2018-01-23T08:47:33

$x = \frac{d - b c}{1 - a b}$ $x=(d-bc)/(1-ab)$ and $y = \frac{c - a d}{1 - a b}$ $y=(c-ad)/(1-ab)$

The best way is to use substitution method. We get $y$ $y$ from first equation and put its value from it in second equation to get $x$ $x$ .,

As $a x + y = c$ $ax+y=c$ , we have $y = c - a x$ $y=c-ax$

putting this value in second equation we get

$x + b (c - a x) = d$ $x+b(c-ax)=d$

or $x + b c - a b x = d$ $x+bc-abx=d$

or $x (1 - a b) = d - b c$ $x(1-ab)=d-bc$

i.e. $x = \frac{d - b c}{1 - a b}$ $x=(d-bc)/(1-ab)$

Hence $y = c - a \times \frac{d - b c}{1 - a b}$ $y=c-axx(d-bc)/(1-ab)$

= $\frac{(c - a b c - a d + a b c)}{1 - a b}$ $((c-abc-ad+abc))/(1-ab)$

i.e. $y = \frac{c - a d}{1 - a b}$ $y=(c-ad)/(1-ab)$

How to solve each of the following pairs of simultaneous equations for x and y? 1) ax+y=candx+by=dax + y = c and x+by=d