Solve by cross multiplication method 5ax …
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Sia ? 6 years, 5 months ago
The systems of equations are:
5ax + 6by = 28 .......................(i)
3ax + 4by = 18 .......................(ii)
From (i) and (ii), we get
a1 = 5a, b1 = 6b, c1 = -(28)
a2 = 3a, b2 = 4b, c2 = -(18)
By cross multiplication method, we get
{tex}\frac{x}{{-108b +112b}} = \frac{{ - y}}{{ - 90a + 84a }} = \frac{1}{{20ab - 18ab}}{/tex}
{tex}\frac{x}{{4b}} = \frac{{ - y}}{{ - 6a}} = \frac{1}{{2ab}}{/tex}
Now, {tex}\frac{x}{{4b}} = \frac{1}{{2ab}} {/tex}
{tex}⇒ x = \frac{2}{a}{/tex}
And, {tex}\frac{{ - y}}{{ - 6a}} = \frac{1}{{2ab}} {/tex}
{tex}⇒ y = \frac{3}{b}{/tex}
Therefore the solution of the given system of equation is {tex}\frac{2}{a}{/tex} and {tex}\frac{3}{b}{/tex}.
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