Alg - Eqns - Simult TYS 1 Solns

Solutions for the simultaneous equations using the equation with 1 in front of one of the variables:

1. y = 2x + 7

y = 4x + 13

Equating the y terms:

2x + 7 = 4x + 13

-6 = 2x

x = -3.

y = 2(-3) + 7

= 1

So x = -3 and y = 1.

2. 2a - b = 6

a + 3b = 10

Re-writing Eqn 1:

b = 2a - 6

Substituting into Eqn 2:

a + 3(2a - 6) = 10

7a = 28

a = 4

b = 2(4) - 6 = 2

So a = 4 and b = 2

3. 2x + y = -2

4x - 3y = 31

Re-writing Eqn 1:

y = -2 - 2x

Substituting into Eqn. 2:

4x - 3(-2 -2x) = 31

10x = 25

x = 2.5

y = -2 -2(2.5) = -7

So x = 2.5 and y = -7

4. 5p - 2q = 8

2p - q = 5

Rewriting Eqn. 2:

q = 2p - 5

Substituting into Eqn 1:

5p -2(2p - 5) = 8

p = -2

q = -9

So p = -2 and q = -9

5. y = x + 1

Equate the two equations:

So x = 6 and y = 7

6. 3m - 2n = 10

5m + n = 8

Rewriting Eqn 2:

n = 8 - 5m

Substituting into Eqn 1:

3m - 2(8-5m) = 10

13m = 26

m = 2

n = 8 - 5(2) = -2

So m = 2 and n = -2

Solutions for the simultaneous equations having equal coefficients for one variable:

7. 2x + y = 1

2x + 3y = -9

Eqn 2 - Eqn 1:
(to avoid the negative sign for y)

2y = -10

y = 5

2x = -4

x = -2

So x = -2 and y = 5

8. 4a - 3b = 11

4a + 2b = 10

Eqn 2 - Eqn 1:
(to avoid the negative sign for b)

5b = -1

b = -0.2

4a = 11 + 3(-0.2) = 10.4

a = 2.6

So a = 2.6 and b = -0.2

9. 5e - 3f = 20

2e + 3f = 15

Adding the eqns:

7e = 35

e = 5

2(5) + 3f = 15

3f = 5

10. 5p - q = 13

2p - q = 7

Eqn 1 - Eqn 2:

3p = 6

p = 2

2(2) - q = 7

q = 4 - 7 = -3

So p = 2 and q = -3

11. 11a + 9b = 34

5a - 9b = -50

Adding the eqns:

16a = -16

a = -1

(substituting into eqn 1 so I can keep the coefficient of b positive)

11(-1) + 9b = 34

9b = 45

b = 5

So a = -1 and b = 5

12. 3y - 4x - 1 = 0

2y + 4x - 14 = 0

Adding the eqns:

5y - 15 = 0

y = 3

(substituting into eqn 2 so I can keep the coefficient of x positive)

2(3) + 4x - 14 = 0

4x = 8

x = 2

So x = 2 and y = 3

Applied questions (mainly of the "break-even" type).

13. After what time and how far from Gunnedah will the brothers meet up to have a coffee and a talk? Chuck drives at 60 kph starting 0 km frpm Gunnedah. So he is 60t km from Gunnedah after t hours. Dave drives at 80 kph from Sydney. So he is 80t km from Sydney after t hours which means he is 430 - 80t from Gunnedah. So distance from Gunnedah after t hours: Chuck: d = 60t Dave: d = 430 - 80 t Equating the equations as the brothers will meet up: 60t = 430 - 80t 140t = 430 t = 3.07 = 3 hours 4 mins. So they have a talk at the rest station after they have driven for 3 hours 4 min which means Chuck has driven about 184 km (3.07 × 60) from Gunnedah. Dave has driven about 246 km (3.07 × 80) from Sydney. One would hope that somewhere each brother would also have had a 20 minute STOP REVIVE SURVIVE pause after 2 hours of driving before meeting up - so then the meeting place would be the same but the elapsed time would be 3 hours 24 mins.

Criterion:

If you attained at least 4 correct in each set and 9 overall, you have made a good start. Re-read the solutions and review at least parts of the video again. Then re-do this test.

If you attained, 10 or 11 correct, you have established a good basis for this topic. It is highly possible your technique for simultaneous equations is sound but that you have made a mistake with your signs when solving equations. Review your work to see if this is the case. Be careful - signs in equations will always be a problem. 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.

If you attained 12 out of 12 - press here.GREAT WORK!!
You should now have lots of confidence on this part of the topic.