unit 6. simultaneous equations - tuprofesorademates - …eso_unit6... · unit 6: simultaneous...

Mathematics 3º ESO. IES Don Bosco (Albacete). European Section 1

Unit 6: SIMULTANEOUS EQUATIONS

This unit will show you how to:

• Solve linear equations graphically

• Solve simultaneous equations by substituting a variable

• Solve simultaneous equations by eliminating a variable

• Write simultaneous equations in order to solve problems

Keywords

Axes Solution

Plot Eliminate

Linear Intersection

Straight line Simultaneous

6.1.- LINEAR EQUATIONS IN TWO VARIABLES. SOLUTIONS

A linear equation in two variables is an equation that can be written in the form ax by c+ = , where a and b are not both 0.

Examples: 2x 5y 7− = , x 3y 0− + =

This form is called the standard form of a linear equation.

The solutions to equations in two variables consist of two values that when substituted into their corresponding variables in the equation, make a true

statement.

We usually write the solutions to equations in two variables in ordered pairs.

Example:

Determine whether each ordered pair is a solution of a given equation.

5x y 7− = ; ( )2,3 , ( )1,5 , ( )1, 12− −

( )x, y 5x y 7− = Solution or not solution?

( )2,3 5 32 7⋅ − = ( )2,3 is a solution to the equation

( )1,5 5 51 7⋅ − ≠ ( )1, 5 is not a solution to the equation

( )1, 12− − ( ) ( )5 721 1−−−⋅ = ( )− −1, 12 is a solution to the equation

2 Mathematics 3º ESO. IES Don Bosco (Albacete). European Section

Note that although you were only given three ordered pairs to check, there is

infinite number of solutions to this equation. It would very cumbersome to find

all of them, so these equations will be solved graphically.

Solving linear equations graphically

The graphs of linear equations in two variables are straight lines.

Example: 2x 5y 7− =

Let’s draw the graph of this equation. (Remember the steps to represent a

function: draw up a table of values, draw x- and y-axes, plot the (x, y) pairs and join them).

In order to draw up the table, it is very convenient rearranging the equation to

make “y” the subject, that is, to get “y” on its own on one side of the equals sign.

2x 72x 5y 7 2x 7 5y y

5

−− = ⇒ − = ⇒ =

x 1 3.5 6 0

y 1− 0 1 1.4−

All the points on this straight line satisfy the equation 2x 5y 7− = , so all the

points in this line are solution of it.

Exercise:

Solve 2x 3y 3− = and 5x 3y 18+ = graphically.

What is the common solution to both equations?


2x 32x 3y 3 2x 3 3y y

3

−− = ⇒ − = ⇒ =

x 0 3 6

y 1− 1 3

18 5x5x 3y 18 3y 18 5x y

3

−+ = ⇒ = − ⇒ =

x 0 3 6

y 6 1 4−

Exercise 1

Are the ordered pairs ( )6, 4 , ( )6, 12 and ( )0, 5 solutions of the equation

4x 3y 12− = ?

Exercise 2

Solve 2x y 6− = and x y 0+ = graphically. What is the common solution to both

equations?

6.2.- SIMULTANEOUS EQUATIONS

Simultaneous equations are a set of equations containing multiple variables. This set is often referred to as a system of equations.

For example:

2x 3y 3

5x 3y 18

− =

+ =

2 2x y 25

x y 1

+ =

− =

2x 4y 5z 3

5x 3y 2z 15

x 6y 2z 3

− + =

+ − =− + + =

Solving simultaneous equations means finding the values of the variables that

make all the equations true at the same time.

The point of intersection of the two

lines is the common solution to both

equations, that is, the point (3, 1).


Example: 2x 3y 3

5x 3y 18

− =

+ =

We saw in a previous exercise that the solution of this system is x 3= and

y 1= .

In this unit you will solve, basically, systems of two linear equations in two

variables.

6.3.- NUMBER OF SOLUTIONS OF TWO-VARIABLE SYSTEMS

When you are solving systems, you are, graphically, finding intersection of lines.

For two-variable systems, there are three possible types of solutions:

Case 1

2x 3y 3

5x 3y 18

− =

+ =

Case 2

2x 3y 15

2x 3y 9

+ =

+ =

Case 3

2x 3y 15

4x 6y 30

+ =

+ =

The two lines cross at exactly one point.

This point is the only solution to the system. These systems are called

independent systems.

Since parallel lines never cross, the

system has no solution. These systems are called inconsistent systems.

The two lines are the same line. Any point of the line is solution to the system. These systems are called dependent

systems.


Exercise 3

Look at the equations of the following systems and determine if they are

independent, inconsistent or dependent systems.

a) 2x y 7

2x y 0

+ =

+ = b)

2x y 7

2x 5y 10

+ =

− + = c)

2x y 7

4x 2y 14

+ =

+ = d)

2x y 7

2x y 1

+ =

− =

6.4.- METHODS TO SOLVE SYSTEMS OF EQUATIONS

Three methods of solving systems of equations will be discussed in this unit: the

graphical method, substitution, and elimination.

The graphical method

The graphical method consists of graphing every equation in the system and then

using the graph to find the coordinates of the point(s) where the graphs

intersect. The point of intersection is the solution.

Examples: Use the graphical method to solve the following systems of equations.

a) x y 2

2x y 3

− = −

− + =

b) 2y x

x y 2

=

− + =

Graph both equations very precisely. If

you don’t graph neatly, your point of

intersection will be way off.

The solution is x 1= − , y 1= . Substitute

these values into both equations to

check the solution.

The graph reveals two solutions: x 1= − ,

y 1= and x 2= , y 4= .

Check both solutions.


As easy as the graphical method is to use, is not very helpful when the solution

is not integers. Imagine trying to use the graphical method to find the solution

if the point of intersection is 37 16

,119 241

.

Graphical method strengths: visualizing possible solutions.

Graphical method weakness: imprecise.

Exercise 4

Use the graphical method to solve the following systems of equations.

a) 2x 3y 0

x 3y 9

− =

+ = b)

x y 1

2x 2y 4

+ =

+ = c)

21y x

2x 2y 2

=

− + =

The substitution method

To solve a system of equations using substitution, first solve one of the

equations for a variable, then substitute it into the other equation(s). Always

substitute into the other equation(s) and always use parentheses.

Example: Use the substitution method to solve the following system of equations.

3x 4y 8

2x 5y 3

+ =

− + =

Step 1. Solve one of the equations for a variable. Let’s solve Eq. 1 for y.

8 3x3x 4y 8 4y 8 3x y

4

−+ = ⇒ = − ⇒ =

Step 2. Substitute this into the other equation and simplify.

2x 5y 3− + =

8 3x2x 5 3

4

−− + =

( )8x 5 8 3x 12− + − =

8x 40 15x 12− + − =

8x 15x 12 40− − = −

23x 28− = −

28 28x

23 23

−= =

−


Step 3. “Back-substitute” this value of x into the equation from step 1 to solve

for y.

288 3

238 3x 25y

4 4 23

− ⋅

− = = =

The solution is 28

x23

= , 25

y23

= . You should put these values into both original

equations to check your work.

Exercise 5

Use the substitution method to solve the following systems of equations.

a) x 3y 0

2x y 5

+ =

+ = − b)

8x 3y 25

x 5y 17

− = −

− = − c)

3x 9y 4

2x 3y 1

+ =

+ =

The elimination method

Elimination consists of adding equations together to eliminate variables.

Sometimes you have to multiply equations by a number before you add them. The

goal is to end up with one equation that has just one variable. Then you can use

back-substitution to solve for the other variable(s).

When using elimination, eliminate one variable at a time. It is also important to

write down “instructions” that indicate how you are manipulating the equations

going from step to step.

Example: Use the elimination method to solve the following system of equations.

5x 3y 7

4x 5y 3

+ = −

+ = −

Let’s get rid of the “y” variable.

Multiplying the first equation by 5 and the second equation by (-3) will make the

“y terms” cancel.

5x 3y 7

4x 5y 3

+ = −

+ = − 5 (eq.1)

( 3) (eq.2)

⋅

− ⋅��

25x 15y 35

12x 15y 9

+ = −

− − =

Now add the two new equations:


25x 15y 35

12x 15y 9

13x 26

x 2

+ = −

− − =

= −

= −

Use back-substitution to find the value of y.

Solving the first equation for y and substituting the value of x:

( )7 5 27 5xy 1

3 3

− − ⋅ −− −= = =

The solution is x 2= − , y 1= .

Exercise 6

Use the elimination method to solve the following systems of equations.

a) 4x 3y 2

2x y 4

− =

+ = − b)

3x y 0

3x y 6

− =

+ = − c)

2x 5y 1

4x 3y 2

+ = −

− = −

Exercise 7

Solve the following systems of equations.

a)

3 y2 x2

3 62 y8 3x

26 9

+−+ =

+− − =

b)

y 1x 11

2 42y 12x 1

12 6

+−+ =

+− − =

Exercise 8

Use the substitution method to solve the following systems of equations.

a) 2

2x y 4

x y 7

+ =

+ = b)

2 2

y x 0

2x y 147

− =

+ = c)

2 2

x y 1

2x y 2

+ =

− =

6.5.- SOLVING PROBLEMS USING SYSTEMS OF EQUATIONS

Remember the steps to solve math word problems:

1. Read the problem entirely (get a feel for the whole problem).

2. List information and the variables you identify (attach units of measure to

the variables).


3. Define what answer you need, as well as its units of measure.

4. Translate the wording into numeric equations.

5. Solve the equation or the system of equations.

6. Check your solution using the problem (not your equations).

Example:

The distance between two railway stations A and B is 255 km. A train leaves

from A to B with a rate of 60 km/h. At the same time another train leaves from

B to A with a rate of 110 km/h. How long do they take to pass each other? What

are the distances that they will have covered?

A 60 km/h Meeting 110 km/h B

x 255 - x

Distance Time Speed

1st train x t 60

2nd train 255 - x t 110

Since distance = speed · time, you can write: 1st train : x 60t

2nd train : 255 x 110t

=

− =

Let’s solve the system of equations: x 60t

255 x 110t

=

− =

Using substitution method:

( )255

255 60t 110t 255 170t t 1.5170

− = ⇒ = ⇒ = =

x 60t 60 1.5 90= = ⋅ =

Solution: The trains take 1h 30 min to pass each other. The first train will have

covered 90 km and the second one, 255 – 90 = 165 km.

Exercise 9

With the wind, an airplane travels 1120 miles

in seven hours. Against the wind, it takes

eight hours. Find the rate of the plane in

still air and the velocity of the wind.


Exercise 10

In a sweet shop, I spend £3.20 on three cans of soft drink and four bars of

chocolate. The next day, I buy a can of soft drink and four bars of chocolate for

£2. How much does each item cost?

Exercise 11

Twice the larger of two numbers is three more than five times the smaller, and

the sum of four times the larger and three times the smaller is 71. What are the

numbers?

Exercise 12

230 students and 29 staff are going on a school trip. They travel by large and

small coaches. The large coaches seat 55 and the small coaches seat 39. If

there are no spare seats and five coaches are to make the journey, how many of

each coach are used?

Exercise 13

Three times the width of a certain rectangle exceeds twice its length by three

inches, and four times its length is twelve more than its perimeter. Find the

dimensions of the rectangle.

Exercise 14

A car and a bus set out at 2 p.m. from the same point, headed in the same

direction. The average speed of the car is 30 mph slower than twice the speed

of the bus. In two hours, the car is 20 miles ahead of the bus. Find the rate of

the car.

Exercise 15

In an isosceles trapezium, the smallest base

is 6 cm. and the length of the largest base

is the same as the two equal sides together.

If the perimeter of the trapezium is 38 cm,

find its dimensions.


Exercise 16

The difference between the two diagonals of a rhombus is 6 cm and its area is

56 cm2. Find the dimensions of the diagonals.

Exercise 17

The sum of the two digits of a number is 7. It you invert the order of the digits,

the number you get is two more than twice the original number. What is the

original number?

Exercise 18

The sides of a triangle are 5 cm, 7cm and 10 cm, respectively. Find the height

relative to the largest side and the area of this triangle.

unit 6. simultaneous equations - tuprofesorademates - …eso_unit6... · unit 6: simultaneous...

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