Transcript 8._Simultaneous_Equations

Mr Barton’s Maths Notes
Algebra
8. Simultaneous Equations
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8. Simultaneous Equations
What are Simultaneous Equations?
Simultaneous Equations are two equations, each containing two unknown letters, and you have to
use both equations, in a clever way, to find the value of your unknown letters!
Key Point: The values you find for your unknown letters must make BOTH equations balance –
and once again this is another Algebra topic where you can check your answers and guarantee
that you have got it right! I told you Algebra wasn’t so bad…
Skills you need to have mastered before we start…
In this section I am going to assume that you are an world expert on the following things:
How to solve equations (see Algebra 4. Solving Equations)
Rules of Algebra (see Algebra 1. Rules of Algebra)
Rules of Negative Numbers (see Number 8. Negative Numbers)
If this is not the case, go back now and have a quick read through!
•
•
•
Please Note: The graphical method for solving simultaneous equations is discussed in Graphs 1.
Straight Line Graphs
How Mr Barton Solves Simultaneous Equations
1. If you need to, re-arrange your equations so they are in the same form
2. Write one equation underneath the other, lining up your unknown letters
3. Choose one of the unknown letters and use your algebra skills to change one or both of the
equations to make sure there are the same number (don’t worry about sign) of your chosen
letter in each equation. Your chosen letter becomes your Key Letter.
4. Put a box around your Key Letters and their sign
5. Follow this rule:
If the signs are the same, subtract the two equations
If the signs are different, then add the two equations
6. If you have done this correctly, your Key Letter should cancel out and you should be left with
just one equation with one unknown
7. Solve this equation to work out the value of the unknown letter
8. Choose one of the original equations and substitute in the answer you found in 7. to work out the
value of the other letter… and try to pick the equation that will make life easy for yourself!
9. Check your answers are correct using the equation you did not choose in 8. !
Example 1
3x  y  19
x  y  9
1. Good news! Our equations are in the same form: some x’s and
some y’s, equal a number!
1
2. Let’s write the second equation underneath the first…
2
3. Okay, so we need to pick either the x’s or the y’s to be our
Key Letter. Well… notice how there are already the same number
of y’s in both equations (there is a disguised 1 in front of both),
so let’s pick the y’s to make life easier for ourselves!
1
4. Put a box around our Key Letters, and their signs:
–
5. The signs of our Key Letters are the same (both +) so we must
Subtract equation 2 from equation 1 .
2 x  10
÷2
8. Use this value in one of the original equations (I’ll chose
to find the value of the other unknown letter:
9. And now we have our two answers:
x  5 y  4
But we may as well check them using equation 2
2
x  y  9
x=5
y=4
5  4  9
1
3x  y  19
x  y  9
2x
6. Our Key Letters have cancelled out, leaving us with a nice
looking equation:
7. Solve it:
2
3x  y  19
x  y  9
1
)
x=5
- 15
 10
x  5
3x  y  19
1 5  y  19
y  4
Example 2
3x  2y  3
2 x  2y  12
1. Good news! Our equations are in the same form: some x’s and
some y’s, equal a number!
1
2. Let’s write the second equation underneath the first…
2
3. Okay, so we need to pick either the x’s or the y’s to be our
Key Letter. Well… notice how there are already the same number
of y’s in both equations (there 2 – don’t worry about the sign!), so
let’s pick the y’s to make life easier for ourselves!
1
4. Put a box around our Key Letters, and their signs:
+
5. The signs of our Key Letters are different (- and +) so we
must Add equation 2 to equation 1.
5x  15
÷5
7. Solve it:
9. And now we have our two answers:
x  3 y  3
But we may as well check them using equation 1
1
3x  2y  3
x=3
y=3
9  6  3
3x  2y  3
2 x  2y  12
5x
6. Our Key Letters have cancelled out, leaving us with a nice
looking equation:
8. Use this value in one of the original equations (I’ll chose 2
to find the value of the other unknown letter:
2
3x  2y  3
2 x  2y  12
2
)
x=3
-6
÷2
 15
x  3
2 x  2y  12
6  2y  12
2y  6
y  3
Example 3
2 x  3y  7
3x  5y  18
1
1. Good news! Our equations are in the same form: some x’s and
some y’s, equal a number!
2
2. Let’s write the second equation underneath the first…
3. Okay, bad news. We don’t have the same number of either
unknown. No problem, though! Why not make the number of x’s
the same by… multiplying 1 by 3 and… multiplying 2 by 2
Note: We could have made the y’s the same if we had liked!
1
x3
2
x2
–
5. The signs of our Key Letters are the same (disguised +) so we
must Subtract equation 2 from equation 1 .
÷ -1
7. Solve it (people seem to mess these ones up…)
9. And now we have our two answers:
But we may as well check them using equation 1
1
2 x  3y  7
x = -19
y = 15
38  45  7
2
)
x  19 y  15
2
6 x  9y  21
6 x  10y  36
 y   15
6. Our Key Letters have cancelled out, leaving us with a nice
looking equation:  y   15
8. Use this value in one of the original equations (I’ll chose 2
to find the value of the other unknown letter:
6 x  9y  21
6 x  10y  36
1
4. Put a box around our Key Letters, and their signs:
2 x  3y  7
3x  5y  18
y = -15
- 75
÷3
y  15
3x  5y  18
3x  75  18
3x   57
x  19
Example 4
7 x  2y   20
3x  6  4 y
1
2nd
1. Bad news! Look at that
equation! Might just have to add 4y
to both sides to sort that mess out!
2. Let’s write the second equation underneath the first…
2
1
3. Okay, bad news. We don’t have the same number of either
unknown. No problem, though! Why not make the number of y’s
the same by… multiplying 1 by 2. The signs will be different,
but who cares?
x2
1
4. Put a box around our Key Letters, and their signs:
+
2
5. The signs of our Key Letters are different (- and +) so we
must Add equation 2 to equation 1 .
7 x  2y   20
3 x  4y  6
14 x  4y   40
14 x  4y   40
3x  4y  6
17 x
  34
6. Our Key Letters have cancelled out, leaving us with a nice
looking equation:
17 x   34
÷ 17
7. Solve it (be careful with negatives!)
8. Use this value in one of the original equations (I’ll chose 2
to find the value of the other unknown letter:
9. And now we have our two answers:
x  2
y  3
But we may as well check them using equation 1
1
7 x  2y   20
x = -2
y=3
2
)
14  6   20
x = -2
+6
÷4
x  2
3 x  4y  6
6  4 y  6
4 y  12
y  3
Quadratics and Simultaneous Equations?
One other thing you might be asked to do is to solve a pair of simultaneous equations where one
of them is a quadratic!
Here is how I do these ones…
How Mr Barton Deal With Quadratics
1. Re-arrange your linear equation so that it is y =
or x =
2. Substitute the linear into the quadratic, being really careful about squares and negatives!
3. Your quadratic should now only have one unknown letter in it (hopefully!). So rearrange it into a
nice form and then solve it by either factorising, or using the Quadratic Formula.
Remember: You will get TWO pairs of answers!
4. Use each of your answers to substitute back into one of the original equations to find TWO
values for the other unknown letter.
5. Check each of these values are correct by subbing into the other original equation
Example 1
y  x2
y  2x  3
1. Good news, the linear equation 2 is already in a very nice form
2. Let’s substitute
2
1
2
into our quadratic expression
Remember: this means that every time we see y in equation 1
we must replace it with 2x + 3
2
3. Okay, so now we have our quadratic, with only one unknown
letter in it (x), instead of two!
– 2x
Let’s re-arrange to make it equal to zero
–3
Now, let’s cross our fingers and hope it factorises… YES!
We solve these in the usual way to get:
x  1 and 3
4. Now it’s time to substitute each of our answers into one of
the original equations (I choose 2
) to find our two values
for y, which gives us our answers:
y  1 and 9
5. But let’s check these by subbing into equation
x = -1
y=1
x=3
y=9
2
1 
 1
y  x2
9 
 3 
y  x
2
2
9  9
2x  3  x2
3  x2  2 x
0  x2  2 x  3
( x  1) ( x  3)  0
x  1
x = -1
1
1  1
1
y  x2
y  2x  3
x=3
x  3
y  2  3
y  1
y  6  3
y  9
Good luck with
your revision!