Question

1. solve: 4r + 5(-8r - 2) = -8r + 5(1 - 5r)

Explanation:

Step1: Expand both sides

First, we expand the expressions on both sides of the equation using the distributive property \(a(b + c)=ab+ac\).
For the left - hand side: \(4r + 5(-8r-2)=4r+5\times(-8r)+5\times(-2)=4r - 40r-10=-36r - 10\)
For the right - hand side: \(-8r + 5(1 - 5r)=-8r+5\times1+5\times(-5r)=-8r + 5-25r=-33r + 5\)

Step2: Move all terms with r to one side and constants to the other

We want to get all the terms with \(r\) on one side and the constant terms on the other side. Let's add \(36r\) to both sides and subtract 5 from both sides.
\(-36r-10+36r - 5=-33r + 5+36r-5\)
Simplifying both sides: \(-15 = 3r\)

Step3: Solve for r

To solve for \(r\), we divide both sides of the equation \(-15 = 3r\) by 3.
\(r=\frac{-15}{3}=-5\)

Answer:

\(r = - 5\)