Question

Which value, when placed in the box, would result in a system of equations with infinitely many solutions?

Answer 1

Answer:

The missing value is 12

Step-by-step explanation:

we know that

If the system of equations has infinitely solutions , then the two equations must be the same

so

$y=-2x+4$ ----> $2x+y=4$  -----> equation A

Multiply equation A by 3 both sides

$3*(2x+y)=4*3$

$6x+3y=12$ ----> equation B

$6x+3y=?$ -----> equation C

equate equation B and equation C

The missing value is 12

Answer 2

Answer:

12

Step-by-step explanation:

We are given that two equations

$y=-2x+4$

$6x+3y=$

We have to find the value in the blank space when we place that value  then system of equations have infinitely many solutions

Equation I can be written as

$2x+y-4=0$

Let a be the value that placed in blank space and system have infinitely many solutions

Then $6x+3y-a=0$

We know that condition of infinite solutions

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

Substitute the values then we get

$\frac{2}{6}=\frac{-4}{-a}$

$\frac{1}{3}=\frac{4}{a}$

$a=4\times 3$

a=12

Hence, when we placed 12 in the box then system of equations would have infinitely many solutions .

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