Question

One equation of a pair of consistent linear equation is 5x + 6y +13 = 0, the second
equation can be
A) 5x +6y + 13 = 0
B) 10x+ 12y + 16= 00
C)x+y+1 =0
D) 15x + 18y + 20 = 0​

Answer 1

Ans:A) 5x+6y+13=0

In option B) 10x+12y+16=0

To get this equation,we have to multiply the first equation with 2

(5x+6y+13=0)×2

we'll get 10x+12y+26=0

but in option B) as we can see that 16 and 26 are completelydifferent values

so this means that it's incorrect

Going to option C) x+y+1=0 this is not possible because 5x+6y+13=0 cannot be shortened to x+y+1= 0

In option d) 15x+18y+20

as we tried in B)

the answer here cannot be the second equation

so A) is the right Answer

Answer 2

Answer:

Option (C) is the correct answer.

Step-by-step explanation:

Consistent linear equation: For a pair of linear equation, there always exists a unique solution.

Consider a pair of linear equations:

$a_1x+b_1y+c_1=0$

$a_2x+b_2y+c_2=0$

Then equations are said to be consistent if,

$\frac{a_1}{a_2}\neq \frac{b_1}{b_2}$

Consider the given equation as follows:

$5x+6y+13=0$

Here, $a_1=5$, $b_1=6$ and $c_1=13$

A) $5x+6y+13=0$

Here, $a_2=5$, $b_2=6$ and $c_2=13$

Then,

$\frac{a_1}{a_2}=\frac{5}{5}=1$ and $\frac{b_1}{b_2}=\frac{6}{6}=1$

⇒ $\frac{a_1}{a_2}=\frac{b_1}{b_2}$, not consistent

Thus, option (A) is not the second equation.

B) $10x+12y+16=0$

Here, $a_3=10$, $b_3=12$ and $c_3=16$

Then,

$\frac{a_1}{a_3}=\frac{5}{10}=\frac{1}{2}$ and $\frac{b_1}{b_3}=\frac{6}{12}=\frac{1}{2}$

⇒ $\frac{a_1}{a_3}=\frac{b_1}{b_3}$, not consistent

Thus, option (B) is not the second equation.

C) $x+y+1=0$

Here, $a_4=1$, $b_4=1$ and $c_4=1$

Then,

$\frac{a_1}{a_4}=\frac{5}{1}=5$ and $\frac{b_1}{b_4}=\frac{6}{1}=6$

⇒ $\frac{a_1}{a_4}\neq \frac{b_1}{b_4}$, a consistent

Thus, option (C) is the second equation.

D) $15x+18y+20=0$

Here, $a_5=15$, $b_5=18$ and $c_5=20$

Then,

$\frac{a_1}{a_5}=\frac{5}{15}=\frac{1}{3}$ and $\frac{b_1}{b_5}=\frac{6}{18}=\frac{1}{3}$

⇒ $\frac{a_1}{a_5}=\frac{b_1}{b_5}$, not consistent

Thus, option (D) is not the second equation.

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