Question

Here is an equation that is true for all values of x: 6(x + 3) = 6x + 18. Londyn saw this equation and says she can tell 18( x + 3) + 32 = 3(6x + 18) + 32 is also true for any value of x. How can she tell? Explain your reasoning.

Answer 1

We need to decide whether the given equations are always or never true for values of x .The given equations are ,

• 
  `x - 12 = x + 1`

solve out for x,

`\longrightarrow x -x =12+1\\`

`\longrightarrow 0 =13\\`

This can never be true. Hence the equation is never true for any values of x.

• 
  `x + (3)/(4) = x - (3)/(4)`

solve out for x,

`\longrightarrow x -x =(3)/(4)+(3)/(4)\\`

`\longrightarrow 0=(3)/(2)`

This can never be true. hence the equation is never true for any values of x.

• 
  `4(x + 3) = 8x + 12 - 4x`

solve out for x,

`\longrightarrow 4x +12=8x+12-4x\\`

`\longrightarrow 4x -8x +4x =12-12\\`

`\longrightarrow 0=0`

hence this equation is true for all values of x.

• 
  `2x - 8 - x = x - 8`

solve out for x,

`\longrightarrow 2x -x -x =8+8\\`

`\longrightarrow 0=0`

hence the equation is true for all values of x.

• 
  `2(x + 5) + 3x = 5(x - 5)`

solve out for x,

`\longrightarrow 2x +10+3x =5x-25\\`

`\longrightarrow 5x -5x =-25-10\\`

`\longrightarrow 0 =-35`

This can never be true. hence the equation is never true for any values of x.

and we are done!