Question

Show that pair of linear equations 3x + 4y =7 and 3/4 x +y =4 has no aolution

Answer 1

3x+4y=7 ..(1) 3/4x+ y=4...(2)=> Multiply the following equation(2) by 4,we get 3x+4y=16 . as you see coefficients of x and y are same so it has no solution.Hope it helps you...

Answer 2

Answer:

We proved that 3x + 4y = 7 and $\frac{3x}{4}$+ y = 4 has no solution.

Step-by-step explanation:

Explanation:

Given in the question that, 3x + 4y = 7 and $\frac{3x}{4}$+ y = 4.

$a_1 = 3 , b _1 = 4\ and\ c_1 = -7$

and ,$a_2 = \frac{3}{4} , b_2 = 1 , c_3 = -4$

• If the two lines are parallel and do not meet, the equations $(a_1x+b_1y+c_1 = 0)$ and $(a_2x+b_2y + c_3 = 0)$ cannot be solved.

• There would not be a solution if $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$.

• An inconsistent pair of linear equations is the name given to this kind of equational system.

Step1:

According to the condition,

$\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$

⇒ $\frac{a_1}{a_2}$ = $\frac{3}{\frac{3}{4} }$ = 4 , $\frac{b_1}{b_2}$ =  $\frac{4}{1}$  = 4 and $\frac{c_1}{c_2} = \frac{-7}{-4}$

Here, we can see that, $\frac{a_1}{a_2} = \frac{b_1}{b_2}\neq \frac{c_1}{c_2}$.

Final answer:

Hence, here we proved that, the given  linear equations has no solution.

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