Consider solving the equation for x.

3(3x + 4) - 5 = ax + b

Which statements are true about the solution to the equation when substituting values for a and b as specified?

Select all that apply.

AIf a=−3 and b=4 , then there is exactly one solution to the equation.
BIf a=−9 and b=13 , then there is no solution to the equation.
CIf a=9 and b=−2 , then there is no solution to the equation.
DIf a=−3 and b=−5 , then there are infinitely many solutions to the equation.
EIf a=9 and b=7 , then there is exactly one solution to the equation.

So let's simplify the equation first.

9x + 12 - 5 = ax + b
9x + 7 = ax + b
9x - ax - b = 7

Alright now plug in the numbers for each option.
I don't feel like doing all of them but I did A (one solution) and B (no solution because the two 9x's cancel out).

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jajumonac jajumonac · Answer 2 · 2020-03-14T21:43:04-04:00

Answer:

A. If a=−3 and b=4 , then there is exactly one solution to the equation.

C. If a=9 and b=−2 , then there is no solution to the equation.

Step-by-step explanation:

The given equation is

$3(3x+4)-5=ax+b$

To solve for $x$, we need to use the distributive property

$3(3x+4)-5=ax+b\\9x+12-5=ax+b\\9x-ax=b-7\\x(9-a)=b-7$

To find the right statement, we need to the given values.

For a=-3 and b=4,

$x(9-a)=b-7\\x(9-(-3))=4-7\\x(9+3)=-3\\12x=-3\\x=\frac{-3}{12} =-\frac{1}{4}$

So, choice A is true, because there's one solution to the equation.

For a=-9 and b=13,

$x(9-a)=b-7\\x(9-(-9))=13-7\\18x=6\\x=\frac{6}{18}=\frac{1}{3}$

The choice B is false.

For a=9 and b=-2

$x(9-a)=b-7\\x(9-9)=-2-7\\0x=-9\\0=-9$

Choice C is correct, because there's no solution, because zero is not equal to -9.

For a=-3 and b=-5

$x(9-a)=b-7\\x(9-(-3))=-5-7\\12x=-12\\x=\frac{-12}{12}=-1$

Choice D is false, becuase there's only one solution.

For a=9 and b=7

$x(9-a)=b-7\\x(9-9)=7-7\\0x=0\\0=0$

Choice E is false, because there are infinite solutions in ths case.

Therefore, the right choices are A and C.