Which values of p and q result in an equation with no solutions? px - 45= qx+75 a. p=-45 and q=-75 b. p=45 and q=-15 c. p=15 and q=15 d. p=15 and q=75

Px - 45 = QX + 75

Px - Qx = 75 + 45 = 120

A. If P = - 45 and Q = - 75, then

- 45x + 75 x = 120

30x = 120 and x = 40

B. If P = 45 and Q = - 15, then

45x - 15x = 120

30x = 120

x = 40

D. If P = 15 and Q = 75, then

15x - 75x = 120

- 60x = 120

x = - 2

C. If P = 15 and Q = 15, then

15x - 15x = 120

0 = 120 which is absurd.

Hence, P = 15 and Q = 15 result in an equation with no solution.

starting with:    

5 + x - 12 = 2x - 7   

5 + x - 12 - x = 2x - 7 - x (subtract x from both sides)   

5 - 12 = x - 7 (combine and cancel x's from both sides)   

5 - 12 + 7 = x - 7 + 7 (add 7 to both sides)   

0 = x (add numbers together on both sides)       

so in order for the equation to be true, x must be 0.   

trying x = -0.5   

5 + x - 12 = 2x - 7 (original equation)   

5 + -0.5 - 12 = 2(-0.5) - 7 (plug in value of x)   

5 + -0.5 - 12 = -1 - 7 (did multiplication)   

-7.5 = -8 (did addition on both sides)       

since -7.5 is obviously not equal to -8, the value of -0.5 for x is wrong.   

trying x = 0   

5 + x - 12 = 2x - 7 (original equation)   

5 + 0 - 12 = 2(0) - 7 (plug in value of x)   

5 + 0 - 12 = 0 - 7 (did multiplication)   

-7 = -7 (did addition on both sides)       

since -7 equals -7, that demonstrates that the value of 0 for x is correct.   

trying x = 1   

5 + x - 12 = 2x - 7 (original equation)   

5 + 1 - 12 = 2(1) - 7 (plug in value of x)   

5 + 1 - 12 = 2 - 7 (did multiplication)   

-6 = -5 (did addition on both sides)       

since -6 is obviously not equal to -5, the value of 1 for x is wrong.