Question

Isaac wants the equation below to have no solution when the missing number is placed in the box.

Which number should he place in the box?

Answer 1

Answer: 4

Step-by-step explanation: To have an equation with no solution you have to get something like 0=1 meaning you have to remove all the variables. In this equation you would distribute to get xy+2y+2x=2x+12+4x where y is the missing number. then combining like terms you get xy+2x+2y=6x+12. Subtracting 2x from both sides you get xy+2y=4x+12. Now to get the variables gone y has to be 4 so that the variables to cancel out. By plugging in 4 you can see this works; 4x+8=4x+12 and by subtracting 4x from both sides you get 8=12 which is not true meaning there is no solution.

Answer 2

`^{\boxed {}} = 4.`

EXPLANATION.

The given equation is

`\boxed {}(x + 2) + 2x = 2(x + 6) + 4x`

Let us expand the right hand side to obtain:

`\boxed {}(x + 2) + 2x = 2x + 12+ 4x`

We group similar terms and keep the expression with the box on the left.

`\boxed {}(x + 2) = 2x - 2x+ 12+ 4x`

Simplify

`\boxed {}(x + 2) = 4x + 12`

We can see that the value of box the will be the equation to have no solution is 4.

Let us substitute 4 for box.

`4(x + 2) = 4x + 12`

`4x + 8 = 4x + 12`

`4x - 4x = 12 - 8`

`0 = 4`

This is not true. Hence the equation has no solution when

`\boxed {} = 4`