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Tiếng Việt

Solving simultaneous linear equations

Lesson 1.2 · 20 min read · Updated

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At the school canteen, Mai buys 22 cups of milk tea and 11 cake for 6565 thousand đồng. Tú buys 11 cup of milk tea and 11 cake for 4040 thousand đồng. How much is a cup of milk tea, and how much is a cake?

Let the price of milk tea be xx and the price of a cake be yy (thousand đồng). Then 2x+y=652x+y=65 and x+y=40x+y=40. In the last lesson we could only check a solution; in this lesson we learn how to find it.

The idea: get rid of one unknown

Explore: comparing the two bills

The two bills differ by exactly 11 cup of milk tea, and by 65−40=2565-40=25 thousand đồng. So a cup of milk tea costs 2525 thousand đồng, and a cake costs 40−25=1540-25=15 thousand đồng.

We removed the unknown yy by subtracting the equations, which left an equation in one unknown. Every method for simultaneous equations follows this idea: transform the equations until one of them has only one unknown. There are two ways to do it: the substitution method and the elimination method.

The substitution method

-1123456-2-112345xyO(3, 1)
Two lines meeting at (3,1)(3, 1)

Not every pair of equations has exactly one solution. When we substitute, both unknowns may disappear.

Geometrically: in a) the lines y=2−xy=2-x and y=2.5−xy=2.5-x are parallel; in b) both equations describe the same line.

The elimination method

When the coefficients of the same unknown in the two equations are equal or opposite, we subtract or add the equations so that this unknown disappears.

If the coefficients are not yet equal or opposite, we multiply first. Remember to multiply both sides, including the right-hand side.

Now we solve the canteen problem with the method we have just learnt.

Key terms

Summary

  • The common idea: transform until you have an equation in one unknown.
  • Substitution: write one unknown in terms of the other and substitute into the other equation.
  • Elimination: make the coefficients of one unknown equal (then subtract) or opposite (then add); multiply both sides when needed.
  • If you reach 0x=c0x=c with c≠0c\neq0: no solution. If you reach 0x=00x=0: infinitely many solutions.

Practice

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

Exercise 1

Solve {x=2yx+y=9\begin{cases} x=2y\\ x+y=9 \end{cases} by substitution.

What you need and how to start

What you need: the substitution method.

Steps:

  1. Substitute x=2yx=2y into the second equation.
  2. Find yy, then xx.
Show solution

2y+y=9  ⟺  y=32y+y=9\iff y=3, so x=6x=6.

Answer. (6,3)(6, 3).

Exercise 2

Solve {2x−y=3x+2y=4\begin{cases} 2x-y=3\\ x+2y=4 \end{cases} by substitution.

What you need and how to start

What you need: it is easiest to make an unknown with coefficient 11 or −1-1 the subject.

Steps:

  1. From the first equation: y=2x−3y=2x-3.
  2. Substitute into the second equation and solve.
Show solution

x+2(2x−3)=4  ⟺  5x=10  ⟺  x=2x+2(2x-3)=4\iff 5x=10\iff x=2, so y=2⋅2−3=1y=2\cdot2-3=1.

Answer. (2,1)(2, 1).

Exercise 3

Solve {x+y=7x−y=1\begin{cases} x+y=7\\ x-y=1 \end{cases} by elimination.

What you need and how to start

What you need: the coefficients of yy are opposite, so add the equations.

Steps:

  1. Add to find xx.
  2. Substitute into x+y=7x+y=7 to find yy.
Show solution

2x=8  ⟺  x=42x=8\iff x=4; 4+y=7  ⟺  y=34+y=7\iff y=3.

Answer. (4,3)(4, 3).

Exercise 4

Solve {3x+4y=52x−3y=−8\begin{cases} 3x+4y=5\\ 2x-3y=-8 \end{cases}

What you need and how to start

What you need: multiply both sides so that the coefficients of one unknown are equal.

Steps:

  1. Multiply the first equation by 22 and the second by 33.
  2. Subtract to remove xx.
Show solution

6x+8y=106x+8y=10 and 6x−9y=−246x-9y=-24. Subtracting: 17y=34  ⟺  y=217y=34\iff y=2. Then 3x+8=5  ⟺  x=−13x+8=5\iff x=-1.

Answer. (−1,2)(-1, 2).

Exercise 5

How many solutions does each pair of equations have? a) {2x−4y=6x−2y=3\begin{cases} 2x-4y=6\\ x-2y=3 \end{cases} b) {x−2y=1−2x+4y=5\begin{cases} x-2y=1\\ -2x+4y=5 \end{cases}

What you need and how to start

What you need: after substituting, 0x=00x=0 means infinitely many solutions and 0x=c0x=c (c≠0c\neq0) means no solution.

Steps:

  1. Write x=2y+…x=2y+\dots from the equation in which xx has coefficient 11.
  2. Substitute into the other equation.
Show solution

a) x=3+2yx=3+2y: 2(3+2y)−4y=6  ⟺  0y=02(3+2y)-4y=6\iff 0y=0. b) x=1+2yx=1+2y: −2(1+2y)+4y=5  ⟺  0y=7-2(1+2y)+4y=5\iff 0y=7.

Answer. a) Infinitely many; b) none.

Exercise 6

Let xx and yy (thousand đồng) be the price of an exercise book and of a pen. 33 books and 22 pens cost 3434 thousand đồng; 22 books and 33 pens cost 3131 thousand đồng. Solve {3x+2y=342x+3y=31\begin{cases} 3x+2y=34\\ 2x+3y=31 \end{cases} to find each price.

What you need and how to start

What you need: the elimination method (or a trick: add the equations, then subtract them).

Steps:

  1. Add: 5x+5y=655x+5y=65, so x+y=13x+y=13.
  2. Subtract: x−y=3x-y=3.
  3. Solve the new pair.
Show solution

x+y=13x+y=13 and x−y=3x-y=3, so 2x=162x=16, x=8x=8, y=5y=5.

Answer. A book costs 88 thousand đồng and a pen 55 thousand đồng.

Exercise 7 Review

Find aa and bb so that the graph of y=ax+by=ax+b passes through the points A(1,3)A(1, 3) and B(2,5)B(2, 5).

What you need and how to start

What you need: the point (x0,y0)(x_0, y_0) is on the graph when y0=ax0+by_0=ax_0+b (Grade 8). Each point gives one equation in the unknowns aa, bb.

Steps:

  1. Substitute the coordinates of AA and BB to get two equations.
  2. Subtract them to find aa.
Show solution

a+b=3a+b=3 and 2a+b=52a+b=5. Subtracting: a=2a=2, so b=1b=1.

Answer. a=2a=2, b=1b=1, that is y=2x+1y=2x+1.

Exercise 8

Consider {x+2y=53x−y=1\begin{cases} x+2y=5\\ 3x-y=1 \end{cases}

a) Solve the equations by substitution.

b) Solve them by elimination and compare the results.

c) Show that the line y=x+1y=x+1 passes through the point where the lines x+2y=5x+2y=5 and 3x−y=13x-y=1 meet.

What you need and how to start

What you need: the solution of the simultaneous equations gives the coordinates of the point where the two lines meet.

Steps:

  1. a) Write x=5−2yx=5-2y.
  2. b) Multiply the second equation by 22 and add.
  3. c) Substitute the coordinates of that point into y=x+1y=x+1.
Show solution

a) 3(5−2y)−y=1  ⟺  15−7y=1  ⟺  y=23(5-2y)-y=1\iff 15-7y=1\iff y=2, x=1x=1.

b) x+2y=5x+2y=5 and 6x−2y=26x-2y=2; adding: 7x=77x=7, x=1x=1, y=2y=2. Both methods give the same result.

c) The lines meet at (1,2)(1, 2), and 2=1+12=1+1.

Answer. a), b) (1,2)(1, 2); c) (1,2)(1, 2) lies on y=x+1y=x+1.

Exercise 9 Challenge

Solve {2(x+y)+3(x−y)=133(x+y)−2(x−y)=13\begin{cases} 2(x+y)+3(x-y)=13\\ 3(x+y)-2(x-y)=13 \end{cases}

What you need and how to start

What you need: you can put u=x+yu=x+y, v=x−yv=x-y to get simpler equations, or expand the brackets and simplify.

Steps:

  1. Solve 2u+3v=132u+3v=13, 3u−2v=133u-2v=13.
  2. From x+y=ux+y=u, x−y=vx-y=v find xx and yy.
Show solution

Multiply the first equation by 22 and the second by 33, then add: 13u=6513u=65, u=5u=5; then 3v=33v=3, v=1v=1. From x+y=5x+y=5, x−y=1x-y=1: x=3x=3, y=2y=2. (Expanding gives 5x−y=135x-y=13, x+5y=13x+5y=13, with the same solution.)

Answer. (3,2)(3, 2).

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 of {x+y=5x−y=3\begin{cases} x+y=5\\ x-y=3 \end{cases} is
Question 2. To eliminate yy from {2x+3y=75x−2y=8\begin{cases} 2x+3y=7\\ 5x-2y=8 \end{cases}, you should
Question 3. The simultaneous equations {x−2y=3−2x+4y=1\begin{cases} x-2y=3\\ -2x+4y=1 \end{cases}
Question 4. 22 bánh mì (Vietnamese baguettes) and 11 carton of milk cost 3131 thousand đồng; 11 bánh mì and 22 cartons of milk cost 2626 thousand đồng. The price of one bánh mì is

Part II. True or false

Question 1. Consider {3x+y=7x−y=1\begin{cases} 3x+y=7\\ x-y=1 \end{cases}
a) Adding the two equations gives 4x=84x=8.
b) The solution is (2,1)(2, 1).
c) From the second equation, y=x+1y=x+1.
d) The solution satisfies x+y=3x+y=3.

Part III. Short answer

Question 1. Solve {2x+y=13x−y=9\begin{cases} 2x+y=1\\ 3x-y=9 \end{cases} and find x+yx+y.

Question 2. The graph of y=ax+by=ax+b passes through (1,4)(1, 4) and (3,0)(3, 0). Find bb.

Show answers and explanations
  1. Part I, question 1: A. Adding: 2x=82x=8, x=4x=4, y=1y=1. B swaps the order; C fits only the first equation; D forgets to divide 2x=82x=8 by 22 (giving x=8x=8, then y=5−8=−3y=5-8=-3).
  2. Part I, question 2: A. This gives 6y6y and −6y-6y, which are opposite, so add: 19x=3819x=38. C subtracts, so yy remains; D eliminates xx, not yy.
  3. Part I, question 3: B. Substitute x=3+2yx=3+2y: −6−4y+4y=1  ⟺  0y=7-6-4y+4y=1\iff 0y=7, no solution. (3,0)(3, 0) fits only the first equation.
  4. Part I, question 4: A. Adding: 3(x+y)=573(x+y)=57, x+y=19x+y=19; subtracting: x−y=5x-y=5; so x=12x=12, y=7y=7. B is the price of milk, C the sum, D the difference.
  5. Part II, question 1: a) True; b) True; c) False; d) True. a) True. b) True: 6+1=76+1=7, 2−1=12-1=1. c) False: y=x−1y=x-1. d) True: 2+1=32+1=3.
  6. Part III, question 1: -1. Adding: 5x=105x=10, x=2x=2, y=1−4=−3y=1-4=-3, so x+y=−1x+y=-1.
  7. Part III, question 2: 6. a+b=4a+b=4, 3a+b=03a+b=0; subtracting: 2a=−42a=-4, a=−2a=-2, b=6b=6.