Exercise 79 Page 145

Practice makes perfect

We can determine how many solutions the equation can have by looking at the highest power of x. Here, that is 1, so the equation has zero, one, or infinitely many solutions. Let's solve it to determine which it is.

4x+3=3x+3

SubEqn

LHS-3=RHS-3

4x=3x

SubEqn

LHS-3x=RHS-3x

x=0

The equation has one solution.

The highest power of x is 1 so this equation has zero, one, or infinitely many solutions.

3(x-4)-x=5+2x

Distr

Distribute 3

3x-12-x=5+2x

SubTerm

Subtract term

2x-12=5+2x

SubEqn

LHS-2x=RHS-2x

-12≠5

- 12 is never equal to 5. That means there is no x for which the original equality holds. Therefore, the equation has no solutions.

Here we have a product that is equal to 0. According to the Zero Product Property, that means at least one of the factors must be zero.

5x-2&=0 or x+4&=0 Each of these equations has one solution, so the original equation has two solutions.

If we can factor the left-hand side we can use the Zero Product Property to determine the number of solutions. To do that, we rewrite -4x as -2x-2x.

x^2-4x+4=0

WriteDiff

Write as a difference

x^2-2x-2x+4=0

FactorOut

Factor out x

x(x-2)-2x+4=0

FactorOut

Factor out 2

x(x-2)-2(x-2)=0

FactorOut

Factor out (x-2)

(x-2)(x-2)=0

ProdToPowTwoFac

a* a=a^2

(x-2)^2=0

The only way the left-hand side can be equal to 0 is if x-2=0. Therefore, this equation has one solution.

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