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Solving Equations
X + 3 = 10
4y = 12
2x + 3 =
13y + 2
12
3x 15 2x + 3 = 11
y – 4 = 7Great Marlow School Mathematics Department
Solving an equation is like keeping a scale balanced.
x 12 Here we can see that x = 12 because the scale is balanced.
Great Marlow School Mathematics Department
x + 3 7
On this scale
x + 3 = 7
The scale is balanced.
But if I take the 3 off…
x +3
77
The scale is not balanced any more.
Great Marlow School Mathematics Department
x7
I can balance it again by taking 3 away from the 7.
4
-3
x
This should be written like this to show all the steps needed to work out the answer.
X + 3 = 7
- 3 = -3
X = 4Great Marlow School Mathematics Department
y - 2 5
On this scale y – 2 = 5
Great Marlow School Mathematics Department
If I add 2 to the -2 it will become 0. Then the “y” will be on its own, but the scale will not be balanced.
Y (– 2 + 2)
5
Remember (-2 + 2 = 0.)
Great Marlow School Mathematics Department
Y – 2 = 5 + 2 = + 2Y = 7
Y (– 2 + 2)
5
How can I balance the scale?
Yes, add 2 to the other side of the scale.
y 7
It should be written like this.
Great Marlow School Mathematics Department
Now you try some. Remember to show all the steps.1) x + 3 = 8 - 3 = - 3 x = 5
2) x + 7 = 15 - 7 = - 7 x = 8
3) Y + 5 = 9 - 5 = - 5 y = 4
4) z + 5 = 12 - 5 = - 5 z = 7
5) a + 4 = 15 - 4 = - 4 a = 11
6) Y - 6 = 14 + 6 = +6 y = 20
7) d -5 = 8 + 5 = + 5 d = 13
8) x - 6 = 15 + 6 = + 6 x = 21
9) Y + 13 = 24 - 13 = - 13 y = 11
Great Marlow School Mathematics Department
Here 3y = 21
Remember that 3y means 3 X y
3y 21
I can’t take the 3 away because it has not been added to the “y”. I must divide the 3y into 3 equal pieces, each piece being 1y.
Great Marlow School Mathematics Department
213y3
What’s wrong?
Correct – it is not balanced.
How can I balance the scale?
That’s right – divide 21 into three equal pieces.
3
y 7
21
Great Marlow School Mathematics Department
It should be written like this.
3y = 213 3 y = 7
Great Marlow School Mathematics Department
How do you think we can solve this equation?
Y2 12
In this equation y divided by 2 equals 12.
Great Marlow School Mathematics Department
Y2 12
Correct! Multiply both sides by 2
X 2 X 2
Remember, if you make a change on one side of the equation, you must make the same change on the other side to keep the equation balanced.
y 24
Great Marlow School Mathematics Department
Now you try some.
1)3 a = 18 4) 2b = 73 3 2 2
a = 6 b = 3.5
2) 5x = 30 5) 7y = 63 5 5 7 7x = 6 y = 9
3) 6z = 42 6) 4d = 48 6 46 4z = 7 d = 12
7) z = 45
z x 5 = 4 x 55
z = 20
8) y = 38
y x 8 = 3 x 88
y = 24Great Marlow School Mathematics Department
2x + 3 19
If we knew what 2x was equal to the we could work out what 1x was equal to.
What do we need to get rid of?
Correct – the +3Great Marlow School Mathematics Department
2x+ 3 19 2x -3
Take 3 away from both sides.
16
Now 2x = 16x 8
Divide both sides by 2.
2x + 3 = 19
-3 = -3
2x = 162 2
X = 8
Any questions?
5x + 3 = 48
-3 = -3
5x = 455 5
X = 9
6x + 9 = 39
-9 = -9
6x = 306 6
X = 5
2x - 3 = 19
+3 = +3
2x = 222 2
X = 11
7x - 6 = 29
+6 = +6
7x = 357 7
X = 5
3x + 5 = 17
-5 = -5
3x = 123 3
X = 4
4x -5 = 19
+5 = +5
4x = 244 4
X = 6
1)
2)
4)
3)
5) 6)
Great Marlow School Mathematics Department
5x + 3
2x + 15
This time there are x’s on both sides of the balance.
Any suggestions?
We could take the 2x off both sides
153x + 3
Now it is just like the previous equations.
5x + 3 = 2x +15
-2x = -2x
3x + 3 = 15
-3 = -3
3x = 1233
X = 4Great Marlow School Mathematics Department
4x + 2 = 2x +16
-2x = -2x
2x + 2 = 16
-2 = -2
2x = 1422
X = 7
6x + 8 = 3x +23
-3x = -3x
3x + 8 = 23
-8 = -8
3x = 1533
X = 5
7x - 10 = 5x +4
-5x = -5x
2x - 10 = 4
+10 = +10
2x = 1422
X = 7
5x + 14 = 7x +3
-5x = -5x
14 = 2x + 3
-3 = -3
11 = 2x22
5.5 = x or x = 5.5
9x - 5 = 2x +16
-2x = -2x
7x - 5 = 16
+5 = +5
7x = 2177
X = 3
3x + 8 = x +15
-x = -x
2x + 8 = 15
-8 = -8
2x = 722
X = 3.5Great Marlow School Mathematics Department
National curriculum reference: A3d Date: June 1995 Paper: 1-----------------------------------------
The perimeter of this triangle is 31 cm.
Work out the value of c.
2c-3+2c+3+4c-5=31
2c+2c+4c-3+3-5=31
8c-5=31
+5=+5
8c = 368 8
c =4.5
Great Marlow School Mathematics Department
Date: June 1997 Paper: 1-----------------------------------------
(a) Solve 3x = 24(b) Solve 18 + 3y = 6 - y. [4]
Date: June 1999 Paper: 1-----------------------------------------(a) Solve the equation 2x = 10.
x = .............. (1 mark)(b) Solve the equation 6y + 1 = 25.
y = ............... (2 marks)(c) Solve the equation 8p - 3 = 3p + 13.
p = .............. (2 marks)
See the next slide for the answers.Great Marlow School Mathematics Department
3x = 243 3
X = 8
18 + 3y = 6 – y
+y = +y
18 + 4y = 6
-18 = -18
4y = -124 4Y = -3
2x = 102 2
6y + 1 = 25
- 1 = -1
6y = 24
X = 5
Y = 46 6
8p – 3 = 3p + 13
-3p = -3p
5p – 3 = 13
+3 = +3
5p = 165 5p = 3.2 or 3 1/5
Great Marlow School Mathematics Department