Dr. J's Maths.com
Where the techniques of Maths
are explained in simple terms.

Algebra - Simultaneous equations - Equating coefficients.
Test Yourself 1 - Solutions.


 

Solve the following equations simultaneously by first equating the coefficients for one variable.

Note if we are given an equation with a variable having a coefficient of 1, we would simply revert to our previous technique.

 

2x + 3y = 19

3x + 2y = 16

To equate the y coefficients multiply Eqn 1 by 2 and Eqn 2 by 3
then subtract Eqn 1 from Eqn 2 :

4x + 6y = 38

9x + 6y = 48

5x = 10

x = 2

2(2) + 3y = 19

3y = 15

y = 5

So x = 2 and y = 5

3x - 5y = 11

2x - 3y = 8

To equate the y coefficients multiply Eqn 1 by 3 and Eqn 2 by 5
then subtract Eqn 1 to Eqn 2 :

9x - 15y = 33

10x - 15y = 40

x = 7

10(7) - 15y = 40

15y = 30

y = 2

So x = 7 and y = 2.

 

12y + 7z - 13 = 0

6y + z + 11 = 0Here we only need to multiply Eqn 2 by 2 to equate the y coefficients. Less work!!!
Of course we could also have just rewritten equation 2 to be z = ... and then substituted. Numbers are getting a bit big but no real problem either way.
Go with your heart.

Eqn 2 multiplied by 2 becomes

12y + 2z + 22 = 0

Subtracting the modified Eqn 2 from Eqn 1 gives

5z = 35

z = 7

6y + 7 + 11 = 0

6y = -18

y = -3

So y = -3 and z = 7

 

7a - 3h = 9

5a + 2h = 23

To equate the h coefficients (which have different signs) multiply Eqn 1 by 2 and Eqn 2 by 3
then add Eqn 1 to Eqn 2.
Substitute the value for a into the original Eqn 2 to use the plus sign in front of h (avoid negatives when you can - they might be dangerous :)

14a - 6h = 18

15a + 6h = 69

29a = 87

a = 3

5(3) + 2h = 23

2h = 8

h = 4

So a = 3 and h = 4.

3m + 4n = -5

4m - 3n = 10

To equate the n coefficients (which have different signs) multiply Eqn 1 by 3 and Eqn 2 by 4
then add Eqn 1 to Eqn 2. Substitute the value for m into the original Eqn 1 to use the plus sign in front of the n.
Remember avoiding negative signs is sometimes a positive thing to do:

9m + 12n = -15

16m - 12n = 40

25m = 25

m = 1

9(1) + 12n = -15

12n = -24

n = -2

So m = 1 and n = -2

3c + 4d = 16

7c - 2d = 60

To equate the d coefficients (which have different signs) multiply Eqn 2 by 2 and then add Eqn 1 to Eqn 2:

14c - 4d = 120

17c = 136

c = 8

3(8) + 4d = 16

4d = -8

d = -2

So c = 8 and d = -2

Criterion:

If you attained at least 4 correct, you have made a good start. Re-read the solutions and review at least parts of Video 3 again. Then re-do this test.

If you attained, 5 correct, you have established a good basis for this topic. Compare your solutions with those above. The point where there is a difference highlights where you are making the mistake. Make a note of mistake in your revision notes so, through review, you will remember not to make that error again.

Again, talk to yourself about what you should be doing at each step so as to keep your concentration high and to ensure you take the right action.