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2º ESO
1
UNIT 7
SIMULTANEOUS EQUATIONS
1. Definitions
A linear equation with two unknowns is an equation with two unknowns which have
both of them degree one, i.e., an equation like
ax + by = c.
Example. 3x + 2y = 5 and  2x + 6y =  9.
A simultaneous linear equation is an equation consisting of several linear equations
considered at the same time.
Another name is system of linear equations.
The simultaneous equations we are going to study are all linear (degree 1).
x  2y  1 
Example.

6x  y  4 
This is the standard form for writing equations when they are part of a system of
equations:

the variables go in order on the left side and the constant term is on the right.
The curly bracket, }, llave, on the right indicates that the two equations are intended to
be solved simultaneously.
Example.
x  y  10

x y 2 
If we consider the equation x + y = 10, we have two unknowns and many pairs of
values of x and y fit the equation:
x = 1 and y = 9,
x = 2 and y = 8,
x = 2.9 and y = 7.1;
even decimal numbers: x = 1.005 and y = 8.995, etc
x  y = 2 is another equation with the same two unknowns. Again, many pairs of values
of x and y fit the equation. For example:
x = 4 and y = 2,
x = 7 and y = 5,
x = 2.9 and y = 0.9, etc.
There is only one pair of values of x and y which fit both equations:
x = 6 and y = 4, because 6+4 = 10 and 6  4 = 2.
Susana López
2º ESO
2
Definition: For a pair of simultaneous equations, a solution is a pair of numbers that
make both equations true when substituted into the equations.
Remark: “a number” can be a negative integer number and also a fraction.
To solve simultaneous equations you need to find values which fit both equations
simultaneously.
Sometimes it is useful to label the equations with capital letters as Equation A and
Equation B.
2. Equivalent equations
Two equations are said to be equivalent if they have the same solutions.
For example, equation 3x = 6 is equivalent to 4x = 8, because their solutions are both
x = 2.
For a pair of simultaneous equations, one of the equations can be substituted by
an equivalent one. The simultaneous equations obtained are said to be equivalent to
the original ones.
There are different ways of getting equivalent equations:
Rule. If you do the same to both sides of an equation, it is still true and the new equation
is equivalent to the original one.
 You can multiply/divide all the terms by the same number.
For example:
2 x  4 y  8 is equivalent to x  2 y  4 , dividing by 2 all the terms.
 You can add/subtract the same number to both members.
For example: subtracting 9 from both members of x + 9 = 1, we get x = 8.
5 x  6 y  10  2 is equivalent to 5 x  6 y  12 , adding 10.
If an equation is multiplied by 1 the result is that all the coefficients change their
sign.
If we substitute one of the equations of a system by an equivalent equation, the new
system is equivalent to the original one: the solutions are still the same.
Exercise. Give an equivalent simultaneous equation to the following, so that they are
written in standard form or in a simpler form.
a)
x  y  10  0

x  4 y

Susana López
b)
x  12  y 

 y  4  x
c)
10 x  5 y  15

9 x  6 y  39 
2º ESO
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3. Methods for solving simultaneous equations
Simultaneous equations can be solved using different methods. All methods are based
on rules for getting equivalent equations.
Several ways to solve simultaneous equations means several ways to have
fun. Think about it  Joy of joys!!
3. 1. The substitution method
1.
2.
3.
4.
5.
Work out one of the unknowns from one equation.
Substitute the expression into the other equation.
Work out this equation with only one unknown.
Get the value of the other unknown.
Check the solution by substituting
( A) x  y  12

( B) x  y  4 
1. We choose x and equation (A). Work out x:
Example.
x = 12 – y
2. Substitute x = 12  y into equation (B): (12  y) – y = 4
ONLY ONE UNKNOWN FROM THIS STEP
3. Work out the y from the equation we have got
12  2y = 4 
12  4 = 2y 
8 = 2y 
4=y

4. Get the value of x:
x = 12 – y = 12 – 4 = 8
x=8
5. Check the solution by substituting into the original equations:
(A) 8 + 4 = 12 OK
(B) 8  4 = 4 OK
Susana López
2º ESO
4
3. 2. Equating unknowns or double substitution method.
1.
2.
3.
4.
5.
Work out one of the unknowns from both equations.
Equate both expressions of the same unknown.
Work out this equation with only one unknown.
Get the value of the other unknown
Check the solution by substituting
Example.
( A) 2 x  y  3

( B) 3x  2 y  13
(equate: igualar)
1. We choose x to be worked out from both equations.
3 y
13  2 y
From (A) we get: x 
. From (B) we have: x 
.
2
3
2. Now we equate both expressions of x:
 3  y 13  2 y

2
3
ONLY ONE UNKNOWN FROM THIS STEP
3. Solve this new equation, where there is not any x:
 3  y 13  2 y

2
3
3( 3  y )  2(13  2 y )
 9  3 y  26  4 y
This gives the solution y = 5
3 y  4 y  26  9
7 y  35
4. Get the value of x substituting the value of y into one of the expressions we have
got in step 1:
3 y 35 2
x

  x =1
2
2
2
5. Check the solution: x =1, y = 5
( A) 2 x  y  3

( B) 3x  2 y  13
Susana López
( A) 2  1  5  3
( B) 3  1  2  5  13
OK
2º ESO
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3. 3. The elimination method = the addition or subtraction method
Both equations must be in the standard form a x + b y = c.
You may have to rearrange the equations you are given: transposing terms.
Adding equations is an allowed operation.
1.
2.
3.
4.
Get the same coefficient for the same unknown, but for the sign.
Add both equations. One unknown will be eliminated.
Solve the remaining equation with one unknown.
Get the value of the other unknown by substituting into one of the original
equations.
5. Check the solution by substituting.
Remember: in step 1,
In general the new coefficient will be the l.c.m of the coefficients of the chosen unknown.
Remember: in step 2,
To eliminate terms with opposite signs, add the equations.
Example.
1.
( A) x  2 y  5

( B) 3x  y  8 
Get the same coefficient for the same unknown. We choose y and we do (B)  2.
(because it has already different sign in (A) and (B))
( A) x  2 y  5 
 (These equations are equivalent to A and B together)
(C ) 6 x  2 y  16
x  2y  5 

6 x  2 y  16

7 x  21 (D)
ONLY ONE UNKNOWN FROM THIS STEP
2.
Add both equations:
3.
Solve the equation: 7 x  21 , x 
4.
Get the value of y by substituting into equation (A) the value of x:
(A) x + 2y = 5  3 + 2y =5  2y = 2  y = 1
5.
Check the solution: (A) 3  2 1  5 OK
Susana López
21
,x=3
7
(B) 3  3  1  8 OK
2º ESO
6
Exercises
Exercise 1. Use the three algebraic methods studied to solve each of the following
simultaneous equations. Check the solutions by substituting (only once in each case).
x  2y  8 
x  y  10 
x  2 y  8
3 x  4 y  24
a)
 b)
 c)
 d)

2 x  y  11
2x  y  8
2 x  y  7
2y  x  2 
Solutions. a) (6, 1). b) 6,4 . c) 2,3 . d) (4, 3).
Exercise 2. Solve:
x  y  6
x  y  8
x  2 y  8
a)
b)
c)



2 x  y  7
x  y  2
y  x  2
3x  2 y  12
x  3 y  6
3 x  4 y  24
d)
f)
 e)


y  x 1 
y  2 x  5
2y  x  2 
Solutions. a) (4, 2). b) (3, 5). c) 2,3 . d) ( 2,3) . e) (3, 1). f) (4, 3).
Problems
1. Billy buys 5 first class stamps and 3 second class stamps at cost of £1.93. Jane buys 3
first class stamps and 5 second class stamps at a cost of £1.83. Calculate the cost of a
first class stamp and the cost of a second class stamp.
Solution: 26 p, 21p.
2. Pam buys 6 pencils and 3 pens for 93 pence. Ray buys 2 pencils and 5 pens for 91
pence. Write down two equations connecting the prices of pencils and pens.
Find the cost of a pencil and the cost of a pen.
Solution: Pencil 8p, pen 15p.
3. Apples are x pence per kg. Oranges are y pence each. 5 kg of apples and 30 oranges
cost £9.00. 10 kg of apples and 15 oranges cost £12.60. Write down two equations
connecting x and y. Find the cost of a kilogram of apples and the cost of an orange.
Solution: Apples £1.08 per kg, oranges 12p each.
4. 10 dozen standard eggs and 5 dozen small eggs cost £13.60. 5 dozen standard eggs
and 8 dozen small eggs cost £11.31. Find the cost of a dozen of standard eggs and the
cost of a dozen of small eggs.
Solution: 95p, 82p.
5. Jenny types at x words per minute. Stuart types at y words per minute. When Jenny
and Stuart both type for 1 minute they type a total of 170 words. When Jenny types for
5 minutes and Stuart types for 3 minutes they type a total of 710 words. Calculate x and
y.
Solution: Jenny 100 words, Stuart 70 words per minute.
6. At a café, John buys 3 coffees and 2 teas for £2.30 and Susan buys 2 coffees and 3
teas for £2.20. Calculate the price of a coffee and the price of a tea.
Solution: Coffee 50p, tea 40p.
7. Charlotte and Brendan are buying presents. Charlotte buys one perfume spray and
one bottle of aftershave for £5. Brendan buys three perfume sprays and two bottles of
aftershave for £12.Find the cost of a perfume spray and the cost of a bottle of
aftershave.
Solution: perfume spray £2 and aftershave £3.
8. The sum of the digits of a two-digit number is 7. When the digits are reversed, the
number is increased by 27. Find the number.
Solution: 25.
Of your interest:
http://www.saab.org/mathdrills/lineq.cgi
http://regentsprep.org/Regents/Math/math-a.cfm
http://regentsprep.org/Regents/Math/syslin/AlgSysAdd.htm
Susana López