Question

The equation 2x+3y=9 and 4x+6y=18 have infinitely many solutions is true or false

Answer 1

Answer:

True..and it's consistent

Answer 2

Answer:

Yes,it is true.

Step-by-step explanation:

Given statement is The equation 2x+3y=9 and 4x+6y=18 have infinitely many solutions.

Now question is this statement true or false.

For a system of equations$a_1x+b_1y+c_1=0$and$a_2x+b_2y+c_2=0$

to have infinitely many solutions, the condition to be satisfied is $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$

For better concept we can take one example.

Let two equations are

$x + 5y - 3 = 0.....(1)$and $3x + 15y - 9 = 0.....(2)$

We are comparing equation (1) with$a_1x+b_1y+c_1=0$

and equation (2) with $a_2x+b_2y+c_2=0$

We get,

$a_1 = 1 \\ a_2 = 3 \\ b_1 = 5 \\ b_2 = 15 \\ c_1 = - 3 \\ c_2 = - 9$

So,

$\frac{a_1}{a_2} = \frac{1}{3} \\ \frac{b_1}{b_2} = \frac{5}{15} = \frac{1}{3} \\ \frac{c_1}{c_2} = \frac{ - 3}{ - 9} = \frac{1}{3}$

So, we can say equation (1) and (2) have infinite solutions.

Now given equations are 2x+3y=9....(3) and 4x+6y=18....(4)

We are comparing equation (3) with

$a_1x+b_1y+c_1=0$and equation (4) with$a_2x+b_2y+c_2=0$

So we get,

$a_1 = 2 \\ a_2 = 4 \\ b_1 = 3 \\ b_2 = 6 \\ c_1 = - 9 \\ c_2 = - 18$

Now,

$\frac{a_1}{a_2} = \frac{2}{4} = \frac{1}{2} \\ \frac{b_1}{b_2} = \frac{3}{6} = \frac{1}{2} \\ \frac{c_1}{c_2} = \frac{ - 9}{ - 18} = \frac{1}{2}$

So equation (3) and equation (4) are satisfying condition

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

Hence given two equations have infinite solutions.