Chapter Four — Linear Equations in One Variable
The Humongous Book of Algebra Problems
62
Note: Problems 4.16–4.17 refer to the equation 19 – (x – 5) = 4.
4.16 Solve the equation by isolating x on the right side of the equal sign.
Apply the distributive property to simplify the left side of the equation.
You are directed to isolate x right of the equal sign, so add x to both sides of the
equation to eliminate variable terms on the left side.
24 = 4 + x
Subtract 4 from both sides of the equation to solve for x.
Note: Problems 4.16–4.17 refer to the equation 19 – (x – 5) = 4.
4.17 Solve the equation by isolating x on the left side of the equal sign and verify
that the solutions to Problems 4.16 and 4.17 are equal.
Simplify the left side of the equation.
You are instructed to isolate x on the left side of the equation, so eliminate the
constant left of the equal sign by subtracting it from both sides of the equation.
The equation is not solved for x, because x has a coefficient of –1. Eliminate the
coefficient by multiplying both sides of the equation by –1.
Problems 4.16 and 4.17 have the same solution: x = 20.
4.18 Solve the equation 6x – 3 = 9x + 2 for x.
Group the x–terms left of the equal sign by subtracting 9x from both sides of
the equation.
Group the constant terms right of the equal sign by adding 3 to both sides of
the equation.
Pretend
the negative
sign outside (x – 5)
is a –1 and multiply
x and –5 by –1.
You could
also divide both
sides by –1.