Express the following linear equations in the form $a x+b y+c=0$ and indicate the values of $a, b$ and $c$ in each case:(i) $2 x+3 y=9.3 \overline{5}$(ii) $x-\frac{y}{5}-10=0$(iii) $-2 x+3 y=6$(iv) $x=3 y$(v) $2 x=-5 y$(vi) $3 x+2=0$(vii) $y-2=0$(viii) $5=2 x$

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To do:

We have to express the given linear equations in the form $ax+by+c=0$ and indicate the values of $a, b$ and $c$ in each of the given cases.

Solution:

(i) Given,

$2x+3y=9.3\overline{5}$

We get,

The linear equation in the form $ax+by+c=0$ as,

$2x+3y-9.3\overline{5}=0$

This implies,

$2x+3y+(-9.3\overline{5})=0$

Comparing, $2x+3y+(-9.3\overline{5})=0$ with $ax+by+c=0$

We get,

$a=2$,

$b=3$ and

$c=-9.3\overline{5}$.

(ii) Given,

$x-\frac{y}{5}-10=0$

This implies,

$x+(-\frac{y}{5})+(-10)=0$

Comparing $x+(-\frac{y}{5})+(-10)=0$ with $ax+by+c=0$

We get,

$a=1$,

$b=\frac{-1}{5}$ and

$c=-10$.

(iii) Given,

$-2x+3y=6$

We get,

The linear equation in the form $ax+by+c=0$ as,

$-2x+3y-6=0$

This implies,

$(-2)x+3y+(-6)=0$

Comparing $(-2)x+3y+(-6)=0$ with $ax+by+c=0$

We get,

$a=-2$,

$b=3$ and

$c=-6$.

(iv) Given,

$x=3y$

We get,

The linear equation in the form $ax+by+c=0$ as,

$x-3y=0$

This implies,

$x+(-3y)+(0)c=0$

Comparing $x+(-3y)+(0)c=0$ with $ax+by+c=0$

We get,

$a=1$,

$b=-3$ and

$c=0$.

(v) Given,

$2x=-5y$

We get,

The linear equation in the form $ax+by+c=0$ as,

$2x+5y=0$

This implies,

$2x+5y+(0)c=0$

Comparing $2x+5y+(0)c=0$ with $ax+by+c=0$

We get,

$a=2$,

$b=5$ and

$c=0$.

(vi) Given,

$3x+2=0$

This implies,

$3x+(0)y+2=0$

Comparing $3x+(0)y+2=0$ with $ax+by+c=0$

We get,

$a=3$,

$b=0$ and

$c=2$.

(vii) Given,

$y-2=0$

This implies,

$(0)x+y+(-2)=0$

Comparing $(0)x+y+(-2)=0$ with $ax+by+c=0$

We get,

$a=0$,

$b=1$ and

$c=-2$.

(viii) Given,

$5=2x$

We get,

The linear equation in the form $ax+by+c=0$ as,

$2x-5=0$

This implies,

$2x+(0)y+(-5)=0$

Comparing $2x+(0)y+(-5)=0$ with $ax+by+c=0$

We get,

$a=2$,

$b=0$ and

$c=-5$.

Updated on 10-Oct-2022 13:40:03