Question

Zahra wants the equation below to have an infinite number of solutions when the missing number is placed in the box.

Answer 1

c = -3

Explanation:

Call the unknown c

c(x-3) +2x = -(x-5) +4

Distribute the negative sign

c(x-3) +2x = -x+5+4

Combine like terms

c(x-3) +2x = -x +9

Subtract 2x from each side

c(x-3) +2x-2x = -x-2x +9

c(x-3) = -3x+9

Factor a -3 from the right side

c(x-3) = -3(x-3)

Divide by (x-3)

c = -3

The right and left hand side need to be equal for there to be infinite solutions, so c = -3

Answer 2

box contains - 3

Explanation:

a(x - 3) + 2x = -(x - 5) +4 Remove the brackets on both sides

ax - 3a + 2x = -x + 5 + 4 Subtract 2x from both sides.

ax - 3a + 2x - 2x = -x - 2x + 9

ax - 3a = - 3x + 9

ax - 3(-3) = - 3x + 9

ax + 9 = - 3x + 9

Now if you make a = -3 then both sides with have - 3 for the x coefficient.

Any number could be put in for x and it will make the answer on both sides equal. Notice you could subtract 9 from both sides and it will not change the condition of the problem.