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2015
Year 10 Core Mathematics
Module 4
Simultaneous Linear Equations
Name: ________________________________________________________
Contents:
Set 1 Graphical Solutions of Simultaneous Linear Equations
Set 2 Solving Simultaneous Linear Equations using Substitution
Set 3 Solving Simultaneous Linear Equations using Elimination
Set 4 Writing Equations from worded statements
Set 5 Applications of Linear Sketch Graphs
SAC Test – Set 1-5 inclusive.
Media Files:
Each set has a video link to Dropbox. You can view these lessons as required.
Maths Online has associated lessons that can be attempted as revision prior to
assessment.
Substitution Method I (MOL 4223)
Elimination Method I (MOL 4220)
Problem Solving using Simultaneous Equations (MOL 4226)
Topic Test (MOL 4227)
Software Files:
Graphmatica 2.4 Free graphing software.
Download from the net or SIMON>Simultaneous Equations>media folder.
Assessment:
An in-class test covering set 1-5 inclusive.
Graphical solution of simultaneous
linear equations
Simultaneous linear equations
• Any two linear graphs will meet at a point, unless they are parallel.
• At this point, the two equations simultaneously share the same x- and y-coordinates.
• This point is referred to as the solution to the two simultaneous linear equations.
• Simultaneous equations can be solved graphically or algebraically.
Graphical solution
• This method involves drawing the graph of each equation on the same set of axes.
• The intersection point is the simultaneous solution to the two equations.
• An accurate solution depends on drawing an accurate graph.
• Graph paper or graphing software can be used.
Use the graph of the given simultaneous equations below to
determine the point of intersection and, hence, the solution
of the simultaneous equations.
x + 2y =4
y = 2x — 3
Point of intersection (2, 1)
Solution: x = 2 and y = I
For the following simultaneous equations, use substitution to check if the given pair of coordinates,
(5, —2), is a solution.
3x — 2y = 19 [1]
4y +
x = —3 [2]
3x — 2y = 19 [1] Check equation [2]:
4y + x = —3 [2] LHS = 4y + x RHS = —3
Check equation [1]: = 4(-2) +5
2y = —8+ 5
LHS = 3x — RHS = 19 =-3
= 3(5) — 2(-2) LHS = RHS
= 15+4
=19 In both cases, LHS = RHS. Therefore, the
LHS = RHS solution set (5, —2) is correct.
Graphical Method Video Link
Set 1
Use the graphs below of the given simultaneous equations to write the point of
intersection and, hence, the solution of the simultaneous equations.
a x+y= b x + y = 2
3
x — y = 1 3x — y = 2
d y +
C y—x=4 2x = 3
3x + 2y = 8 2y + x = 0
f 2y — 4x = 5
e y — 3x = 2
4y+ 2x=5
x — y = 2
2 For the following simultaneous equations, use substitution to check if the given pair of
coordinates is a solution.
(7,5) 3x + 2y = 31 (3,7) y — x = 4
a b
2x+3y= 28 2y+x= 17
C (9,1) x + 3y = 12 d (2,5) x+y= 7
5x — 2y = 43 2x+ 3y= 18
e (4, —3) y = 3x — 15 f (6, —2) x — 2y = 2
4x + 7y = —5 3x+y= 16
g (4, —2) 2x + y = 6 h (5,1) y — 5x = —24
x — 3y = 8 3y + 4x = 23
i (-2, —5) 3x — 2y = —4 j (-3, —1) y —x = 2
2x — 3y = 11 2y — 3x = 7
3 Solve each of the following pairs of simultaneous equations using a graphical method.
a x x+2y=10
+ y = 5 b
2x + y = 8 3x+y=15
c d
2x+3y=6 x-3y=-8
2x — y = —10 2x + y = —2
6x+5y=12 f
• y+2x=6
5x + 3y = 10 2y + 3x = 9
Answers
—
Graphical solution of simultaneous
linear equations
c
1 a (2,1) b (1,1) (0,4)
d
(2, —1) e f (-0.5, 1.5)
(-2, —4)
2 a No b Yes C Yes d No
g
e Yes f No No h Yes
i No j Yes
3 a (3, 2) b (4, 3) c (-3, 4) d (-2, 2)
e (2,0) f (3,0)
Solving simultaneous linear
equations using substitution
• There are two algebraic methods which can be used to solve simultaneous equations.
• They are the substitution method and the elimination method.
Substitution method
• This method is particularly useful when one (or both) of the equations is in a form where
one of the two variables is the subject.
• This variable is then substituted into the other equation, producing a third equation with only
one variable.
• This third equation can then be used to determine the value of the variable.
Solve the following simultaneous equations using the substitution method.
y = 2x — 1 and 3x + 4y = 29
V = 2x — 1
3x + 4y = 29
Substituting (2x — 1) into [2]:
3x + 4(2x — 1) = 29
3x + 8x — 4 = 29 [3]
1 lx — 4 = 29
1 lx = 33
x = 3
Substituting x = 3 into [1]:
y = 2(3) — 1
= 6
— 1
=5
Solution: x = 3, y = 5 or (3, 5)
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