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) \( 5=2 x \)
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
$c=-5$.
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