Question

how many solutions exist for the given equation?
0.75(x + 40) = 0.35(x + 20) + 0.35(x + 20)
one
infinitely many
two
zero

Explanation:

Step1: Simplify right - hand side

First, we simplify the right - hand side of the equation \(0.75(x + 40)=0.35(x + 20)+0.35(x + 20)\). Since \(0.35(x + 20)+0.35(x + 20)=0.7(x + 20)\) (by combining like terms, \(0.35+0.35 = 0.7\)).

Step2: Expand both sides

Expand the left - hand side: \(0.75(x + 40)=0.75x+0.75\times40=0.75x + 30\).
Expand the right - hand side: \(0.7(x + 20)=0.7x+0.7\times20 = 0.7x+14\).
Now our equation is \(0.75x + 30=0.7x + 14\).

Step3: Solve for x

Subtract \(0.7x\) from both sides: \(0.75x-0.7x + 30=0.7x-0.7x + 14\), which simplifies to \(0.05x+30 = 14\).
Then subtract 30 from both sides: \(0.05x+30 - 30=14 - 30\), so \(0.05x=-16\).
Divide both sides by \(0.05\): \(x=\frac{-16}{0.05}=- 320\).
Since we have found a unique value for \(x\) that satisfies the equation, the equation has one solution.

Answer:

one