LemonmathSelf-study sheet

Grade 9 · Lesson 2.1

Equations reducible to linear equations in one unknown

Name: Date: Time spent: min

Goals

Solve equations in factorised form $(ax+b)(cx+d)=0$ , and get an equation into this form by factorising.
Find the values of the unknown for which an equation with the unknown in the denominator is defined.
Solve equations with the unknown in the denominator and reject any value that makes a denominator zero.

Remember

$A\cdot B=0\iff A=0$ or $B=0$ . To get a product: move everything to one side and factorise.
Never divide both sides by an expression containing the unknown.
Unknown in the denominator: find the excluded values, use a common denominator and clear the fractions, solve, then reject any excluded value.
An equation may have no solution if every value found is rejected.

Exercises

Exercise 1

Solve the equation $(x-5)(3x+6)=0$ .

Exercise 2

Solve the equation $2x^2+7x=0$ .

Exercise 3

Solve the equation $(2x-1)^2-(x+4)^2=0$ .

Exercise 4

Solve the equation $\dfrac{3}{x-2}=\dfrac{5}{x+2}$ .

Exercise 5

Solve the equation $\dfrac{x+1}{x-3}=2+\dfrac{4}{x-3}$ .

Exercise 6

Solve the equation $\dfrac{2}{x+1}-\dfrac{1}{x-1}=\dfrac{3x-11}{x^2-1}$ .

Exercise 7

With $120$ thousand đồng, Linh can buy as many type I exercise books as she can buy type II books with $90$ thousand đồng. A type I book costs $2$ thousand đồng more than a type II book. Find the price of a type II book.

Exercise 8

Consider the expression $A=\dfrac{x}{x+3}+\dfrac{2}{x-3}-\dfrac{12}{x^2-9}$ .

a) Find the values of $x$ for which $A$ is defined.

b) Simplify $A$ .

c) Find $x$ such that $A=\dfrac12$ .

Exercise 9 Challenge

Solve the equation $x^3+2x^2-x-2=0$ .

Mini test

15 minutes, without looking back at the lesson. Part I: circle one letter; Part II: write T or F for each statement; Part III: write the answer in the box.

Part I. Multiple choice

Question 1. The solution set of the equation

(x-2)(3x+9)=0

S=\{2, -3\}

.B.

S=\{-2, 3\}

.C.

S=\{2, 3\}

.D.

S=\{2, -9\}

Question 2. The equation

\dfrac{x}{x+1}+\dfrac{3}{x-4}=2

is defined when

x\neq-1

and

x\neq4

.B.

x\neq1

and

x\neq-4

.C.

x\neq-1

.D.

x\neq4

Question 3. The solution set of the equation

x^2=5x

S=\{5\}

.B.

S=\{0, 5\}

.C.

S=\{0\}

.D.

S=\{0, -5\}

Question 4. The equation

\dfrac{x+1}{x-2}=\dfrac{3}{x-2}

A. has the solution

x=2

.B. has the solution

x=-2

.C. has no solution.D. has the solution

x=4

Part II. True or false

Question 1. Consider the equation

\dfrac{4}{x+1}=\dfrac{3}{x-2}

a) The equation is defined for $x\neq-1$ and $x\neq2$ .	T / F
b) Clearing fractions gives $4(x-2)=3(x+1)$ .	T / F
c) The equation has the solution $x=-11$ .	T / F
d) The equation has two solutions.	T / F

Part III. Short answer

Question 1. Find the sum of the solutions of the equation

(2x-8)(x+7)=0

Question 2. Solve the equation

\dfrac{5}{x+2}=\dfrac{3}{x-2}

Score: ........ / 10

Hints (only when stuck)

Exercise 1.

What you need: $A\cdot B=0\iff A=0$ or $B=0$ .

Steps:

Set $x-5=0$ .
Set $3x+6=0$ .

Exercise 2.

What you need: take out a common factor to get an equation in factorised form.

Steps:

Write $2x^2+7x=x(2x+7)$ .
Solve each linear equation.

Exercise 3.

What you need: the identity $A^2-B^2=(A-B)(A+B)$ .

Steps:

Factorise the left-hand side as the product of $(2x-1)-(x+4)$ and $(2x-1)+(x+4)$ .
Simplify each factor and solve.

Exercise 4.

What you need: the steps for an equation with the unknown in the denominator.

Steps:

The equation is defined for $x\neq2$ , $x\neq-2$ .
Multiply both sides by $(x-2)(x+2)$ .
Solve and check against the excluded values.

Exercise 5.

What you need: when clearing fractions, multiply every term by the common denominator.

Steps:

The equation is defined for $x\neq3$ .
Multiply both sides by $x-3$ .
Check against the excluded value.

Exercise 6.

What you need: $x^2-1=(x-1)(x+1)$ is the common denominator.

Steps:

The equation is defined for $x\neq1$ , $x\neq-1$ .
Multiply both sides by $(x-1)(x+1)$ .
Solve and check against the excluded values.

Exercise 7.

What you need: number of books $=$ money $\div$ price of one book.

Steps:

Let the price of a type II book be $x$ thousand đồng ( $x>0$ ), so type I costs $x+2$ .
Form the equation $\dfrac{120}{x+2}=\dfrac{90}{x}$ and solve it.

Exercise 8.

What you need: $x^2-9=(x-3)(x+3)$ ; factorise $x^2-x-6$ by splitting $-x=-3x+2x$ .

Steps:

a) Every denominator is non-zero.
b) Use the common denominator $(x-3)(x+3)$ and simplify.
c) Solve $\dfrac{x+2}{x+3}=\dfrac12$ and check against the excluded values.

Exercise 9.

What you need: group terms to find a common factor; an equation in factorised form can have three factors.

Steps:

Group as $(x^3+2x^2)-(x+2)$ .
Take out the common factor $x+2$ , then use $x^2-1=(x-1)(x+1)$ .

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