Question

HELP!! ASAP!!!!! Which system of equations below has infinitely many solutions?

Answer 1

Answer:

The correct option is B. $y=-3x+4 \ \text{and}\ \ 3y=-9x+12$

Step-by-step  explanation:

Consider the two linear equation:

$a_1x+b_1y+c_1=0\ \text{and}\ \ a_2x+b_2y+c_2=0$

The pair of linear equation has infinitely many solutions when:

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

Now consider the option.

Option a) $y=-3x+4 \ \text{and}\ \ y=-3x-4$

The above equation can be written as:

$y+3x-4=0 \ \text{and}\ \ y+3x+4=0$

Therefore,

$\frac{1}{1}=\frac{3}{3}\neq\frac{-4}{4}$

Thus, this pair has no solution.

Now consider the option B. $y=-3x+4 \ \text{and}\ \ 3y=-9x+12$

The above equation can be written as:

$y+3x-4=0 \ \text{and}\ \ 3y+9x-12=0$

Therefore,

$\frac{1}{3}=\frac{3}{9}=\frac{-4}{-12}$

Thus, this pair has infinitely many solution.

Hence, the correct option is B. $y=-3x+4 \ \text{and}\ \ 3y=-9x+12$

Answer 2

Hello,

Answer B since
(y=-3x+4)* 3==> 3y=-9x+12 which is the second equation ==>infinity many solutions.