Question

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

Explanation:

Step1: Simplify the right - hand side

First, we notice that the right - hand side has two identical terms \(0.35(x + 20)\). We can combine them. Using the distributive property \(a\times c+b\times c=(a + b)\times c\), here \(a = 0.35\), \(b=0.35\) and \(c=(x + 20)\). So \(0.35(x + 20)+0.35(x + 20)=(0.35 + 0.35)(x + 20)=0.7(x + 20)\)

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: \(0.7(x + 20)=0.7x+0.7\times20=0.7x + 14\)

Step3: Solve for x

Now our equation is \(0.75x+30 = 0.7x + 14\)
Subtract \(0.7x\) from both sides: \(0.75x-0.7x+30=0.7x - 0.7x+14\), which simplifies to \(0.05x+30 = 14\)
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 can find a unique value of \(x\) that satisfies the equation, the equation has one solution.

Answer:

one