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Tiếng Việt

Equations reducible to linear equations in one unknown

Lesson 2.1 · 20 min read · Updated

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An solves the equation x2=3xx^2=3x like this: "Divide both sides by xx, so x=3x=3." Bình says: "You have lost a solution." Substituting x=0x=0 gives 02=3⋅00^2=3\cdot0, so 00 really is a solution too. Where did An go wrong, and how can we make sure no solution is lost?

The equation x2=3xx^2=3x is not linear, but we can reduce it to linear equations. This lesson covers two kinds of equation like this.

Equations in factorised form

Explore: when is a product equal to 00?

We have 5⋅0=05\cdot0=0 and 0⋅(−7)=00\cdot(-7)=0, but 2⋅3=6≠02\cdot3=6\neq0. A product is 00 only when at least one of its factors is 00. This is the key to solving an equation whose left-hand side is a product of factors and whose right-hand side is 00 (Vietnamese textbooks call it a product equation).

When the equation is not yet in factorised form, we move every term to the left-hand side and factorise (as in Grade 8).

Equations with the unknown in the denominator

Hoà and Bình set off on their bikes at the same time. In the same length of time Hoà rides 1212 km and Bình rides 1818 km, and Bình is 44 km/h faster than Hoà. If Hoà's speed is xx (km/h), equal times give the equation 12x=18x+4.\frac{12}{x}=\frac{18}{x+4}. The unknown is in the denominator. A fraction only makes sense when its denominator is not 00, so first we must find which values of the unknown are allowed.

Checking for excluded values is not a formality: sometimes the equation obtained after clearing fractions gives a value that makes a denominator 00.

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Excluded values and the solution on a number line

Key terms

Summary

  • A⋅B=0  ⟺  A=0A\cdot B=0\iff A=0 or B=0B=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.

Practice

Each exercise comes with what you need and some steps to start. Try it yourself first, then open the solution.

Exercise 1

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

What you need and how to start

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

Steps:

  1. Set x−5=0x-5=0.
  2. Set 3x+6=03x+6=0.
Show solution

x−5=0  ⟺  x=5x-5=0\iff x=5; 3x+6=0  ⟺  x=−23x+6=0\iff x=-2.

Answer. x=5x=5 or x=−2x=-2.

Exercise 2

Solve the equation 2x2+7x=02x^2+7x=0.

What you need and how to start

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

Steps:

  1. Write 2x2+7x=x(2x+7)2x^2+7x=x(2x+7).
  2. Solve each linear equation.
Show solution

x(2x+7)=0  ⟺  x=0x(2x+7)=0\iff x=0 or 2x+7=0  ⟺  x=−722x+7=0\iff x=-\dfrac72.

Answer. x=0x=0 or x=−72x=-\dfrac72.

Exercise 3

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

What you need and how to start

What you need: the identity A2−B2=(A−B)(A+B)A^2-B^2=(A-B)(A+B).

Steps:

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

(2x−1−x−4)(2x−1+x+4)=0  ⟺  (x−5)(3x+3)=0  ⟺  x=5(2x-1-x-4)(2x-1+x+4)=0\iff(x-5)(3x+3)=0\iff x=5 or x=−1x=-1.

Answer. x=5x=5 or x=−1x=-1.

Exercise 4

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

What you need and how to start

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

Steps:

  1. The equation is defined for x≠2x\neq2, x≠−2x\neq-2.
  2. Multiply both sides by (x−2)(x+2)(x-2)(x+2).
  3. Solve and check against the excluded values.
Show solution

3(x+2)=5(x−2)  ⟺  3x+6=5x−10  ⟺  2x=16  ⟺  x=83(x+2)=5(x-2)\iff 3x+6=5x-10\iff 2x=16\iff x=8 (not an excluded value).

Answer. x=8x=8.

Exercise 5

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

What you need and how to start

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

Steps:

  1. The equation is defined for x≠3x\neq3.
  2. Multiply both sides by x−3x-3.
  3. Check against the excluded value.
Show solution

x+1=2(x−3)+4  ⟺  x+1=2x−2  ⟺  x=3x+1=2(x-3)+4\iff x+1=2x-2\iff x=3. This is the excluded value x=3x=3, so it is rejected.

Answer. The equation has no solution.

Exercise 6

Solve the equation 2x+1−1x−1=3x−11x2−1\dfrac{2}{x+1}-\dfrac{1}{x-1}=\dfrac{3x-11}{x^2-1}.

What you need and how to start

What you need: x2−1=(x−1)(x+1)x^2-1=(x-1)(x+1) is the common denominator.

Steps:

  1. The equation is defined for x≠1x\neq1, x≠−1x\neq-1.
  2. Multiply both sides by (x−1)(x+1)(x-1)(x+1).
  3. Solve and check against the excluded values.
Show solution

2(x−1)−(x+1)=3x−11  ⟺  x−3=3x−11  ⟺  2x=8  ⟺  x=42(x-1)-(x+1)=3x-11\iff x-3=3x-11\iff 2x=8\iff x=4 (not an excluded value).

Answer. x=4x=4.

Exercise 7

With 120120 thousand đồng, Linh can buy as many type I exercise books as she can buy type II books with 9090 thousand đồng. A type I book costs 22 thousand đồng more than a type II book. Find the price of a type II book.

What you need and how to start

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

Steps:

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

A price is positive, so x>0x>0. 120x=90(x+2)  ⟺  30x=180  ⟺  x=6120x=90(x+2)\iff 30x=180\iff x=6. Check: 120÷8=15120\div8=15 and 90÷6=1590\div6=15.

Answer. 66 thousand đồng.

Exercise 8

Consider the expression A=xx+3+2x−3−12x2−9A=\dfrac{x}{x+3}+\dfrac{2}{x-3}-\dfrac{12}{x^2-9}.

a) Find the values of xx for which AA is defined.

b) Simplify AA.

c) Find xx such that A=12A=\dfrac12.

What you need and how to start

What you need: x2−9=(x−3)(x+3)x^2-9=(x-3)(x+3); factorise x2−x−6x^2-x-6 by splitting −x=−3x+2x-x=-3x+2x.

Steps:

  1. a) Every denominator is non-zero.
  2. b) Use the common denominator (x−3)(x+3)(x-3)(x+3) and simplify.
  3. c) Solve x+2x+3=12\dfrac{x+2}{x+3}=\dfrac12 and check against the excluded values.
Show solution

a) x≠3x\neq3 and x≠−3x\neq-3.

b) A=x(x−3)+2(x+3)−12(x−3)(x+3)=x2−x−6(x−3)(x+3)=(x−3)(x+2)(x−3)(x+3)=x+2x+3A=\dfrac{x(x-3)+2(x+3)-12}{(x-3)(x+3)}=\dfrac{x^2-x-6}{(x-3)(x+3)}=\dfrac{(x-3)(x+2)}{(x-3)(x+3)}=\dfrac{x+2}{x+3}.

c) 2(x+2)=x+3  ⟺  x=−12(x+2)=x+3\iff x=-1 (not an excluded value).

Answer. a) x≠±3x\neq\pm3; b) A=x+2x+3A=\dfrac{x+2}{x+3}; c) x=−1x=-1.

Exercise 9 Challenge

Solve the equation x3+2x2−x−2=0x^3+2x^2-x-2=0.

What you need and how to start

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

Steps:

  1. Group as (x3+2x2)−(x+2)(x^3+2x^2)-(x+2).
  2. Take out the common factor x+2x+2, then use x2−1=(x−1)(x+1)x^2-1=(x-1)(x+1).
Show solution

x2(x+2)−(x+2)=0  ⟺  (x+2)(x2−1)=0  ⟺  (x+2)(x−1)(x+1)=0x^2(x+2)-(x+2)=0\iff(x+2)(x^2-1)=0\iff(x+2)(x-1)(x+1)=0.

Answer. x=−2x=-2, x=1x=1 or x=−1x=-1.

Mini test

Mini test

Marked like the Vietnamese graduation exam: Part I 0.25 points per question; Part II up to 1 point per question (1, 2, 3, 4 correct statements give 0.1, 0.25, 0.5, 1 point); Part III 0.5 points per question.

Part I. Multiple choice

Question 1. The solution set of the equation (x−2)(3x+9)=0(x-2)(3x+9)=0 is
Question 2. The equation xx+1+3x−4=2\dfrac{x}{x+1}+\dfrac{3}{x-4}=2 is defined when
Question 3. The solution set of the equation x2=5xx^2=5x is
Question 4. The equation x+1x−2=3x−2\dfrac{x+1}{x-2}=\dfrac{3}{x-2}

Part II. True or false

Question 1. Consider the equation 4x+1=3x−2\dfrac{4}{x+1}=\dfrac{3}{x-2}.
a) The equation is defined for x≠−1x\neq-1 and x≠2x\neq2.
b) Clearing fractions gives 4(x−2)=3(x+1)4(x-2)=3(x+1).
c) The equation has the solution x=−11x=-11.
d) The equation has two solutions.

Part III. Short answer

Question 1. Find the sum of the solutions of the equation (2x−8)(x+7)=0(2x-8)(x+7)=0.

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

Show answers and explanations
  1. Part I, question 1: A. x−2=0  ⟺  x=2x-2=0\iff x=2; 3x+9=0  ⟺  x=−33x+9=0\iff x=-3. B has both signs wrong; D forgets to divide by 33.
  2. Part I, question 2: A. Both denominators x+1x+1 and x−4x-4 must be non-zero. B has the wrong signs; C and D look at only one denominator.
  3. Part I, question 3: B. x(x−5)=0x(x-5)=0. Option A is what you get by dividing both sides by xx and losing the solution 00.
  4. Part I, question 4: C. Clearing fractions gives x+1=3x+1=3, x=2x=2, but x=2x=2 makes the denominator 00, so it is rejected. Option A forgets to reject the excluded value; D moves the 11 across without changing its sign (x=3+1x=3+1).
  5. Part II, question 1: a) True; b) True; c) False; d) False. a) True. b) True. c) False: 4x−8=3x+3  ⟺  x=114x-8=3x+3\iff x=11. d) False: there is only one solution, x=11x=11.
  6. Part III, question 1: -3. The two solutions are 44 and −7-7, and their sum is −3-3.
  7. Part III, question 2: 8. Defined for x≠±2x\neq\pm2. 5(x−2)=3(x+2)  ⟺  2x=16  ⟺  x=85(x-2)=3(x+2)\iff 2x=16\iff x=8 (not an excluded value).