Simulatenous Equations Help?

There are two ways you can do these sorts of problems. You can use substitution (where you solve one variable in terms of the other and subtitute it into the other equation), or you can use equation subtraction, where you multiply one of the equations to match up a variable and eliminate it. I like the second one, so:

1)

10p + 12q = 16

9p + 12q = 12 (multiplied the second equation by 3 to match up the q's)

----------------------- (subtraction the bottom equation from the top equation)

p = 4

3p + 4q = 4

3(4) + 4q = 4

12 + 4q = 4

4q = -8

q = -2

2)

8s + 4t = 20

6s + 4t = 22 (multipled the bottom equation by 2 to match the t's)

------------------ (subtraction)

2s = -2

s = -1

3s + 2t = 11

3(-1) + 2t = 11

-3 + 2t = 11

2t = 14

t = 7

3)

15a + 10b = 40 (multiplied the top equation by 5 to match the a's)

15a + 3b = -30

----------------------- (subtraction)

7b = 70

b = 10

15a + 3b = -30

15a + 3(10) = -30

15a + 30 = -30

15a = -60

a = -4

Good luck!

The idea is to eliminate one variable, form a new equation having only the other variable which you can then solve for that variable, and then use that value in one (or both) of the original two equations to solve for the other variable; thus:

10p + 12 q = 16

3p + 4q = 4

The simplest variable to eliminate is the q as you simply have to multiply the second equation by -3 and add:

10p + 12q = 16

-9p - 12q = -12

----------------------

p=4

10(4) + 12q = 16

40 + 12q = 16

12q = -24

q = -2

You can check (if you're interested, by plugging both values into the second equation): 3(4) + 4(-2) = 12 - 8 = 4

Sometimes you have to work a little harder to eliminate a variable in the first step. But your examples are all of the easier sort:

8s+ 4t =20

3s + 2t = 11 multiply by (-2)

8s + 4t = 20

-6s -4t = -22 add

-----------------

2s = -2

s = -1

8(-1) + 4t = 20

-8 + 4t = 20

4t = 28

t = 7

3(-1) + 2(7) = -3 + 14 = 11

3a + 2b = 8

15a + 3b = -30 multiply the first equation by -5

-15a -10b = -40

15a + 3b = -30 add

---------------------

-7b = -70

b = 10

3a + 2(10) = 8

3a + 20 = 8

3a = -12

a = -4

15(-4) + 3(10) = -60 + 30 = -30

Hope you get it.

For the first problem, we're going to multiply the bottom equation by -3

10p+12q=16

-3(3p+4q=4)

10p+12q=16

-9p-12q=-12

p=4. Now, we substitute 4 back into one of the equations and solve for q.

3(4)+4q=4

12+4q=4

-12 -12

4q=-8

/4 /4

q=-2

The same procedures apply for the other problems. Keep that in mind as you continue.

1) because there are 2 unknowns you need to eliminate one.

3p + 4q = 4 multiply the equation by 3 you get 9p + 12q = 12

minus the equations

(10p + 12q = 16) - (9p +12q =12)

p = 4

substitute p = 4

q = -2

10p + 12q = 16 ------ (1)

3p + 4q = 4 ------ (2)

here you have eliminate either p or q you can see12q in eq. 1 and 4q in eq. 2

if you multiply eq. 2 by 3 you will get 12q term in new equation

eq. 2 × 3

9p + 12q = 12 ----- (3)

now there are +12q and + 12q in equations (1) AND (3)

to cancel this you have to subtract the equations

10p + 12q = 16 ------ (1)

9p + 12q = 12 ----- (3)

eq 1 – eq. 3

p = 4

substitute value pf p in any one of the equations say in (2)

3p + 4q = 4 ------ (2)

12 + 4q = 4

4q = – 8

q = –2

8s+4t=20

3s+2t=11 here multiply by 2 and the subtract

3a+2b=8 ---------- multiply by 3

15a+3b=-30 ---------- multiply by 2

then subtract

-----

When there is a system of simultaneous equations with TWO equations on TWO variables, the most general method of solution is as follows.

Step 1.

Select any one of the variable. ( This can be done based on your choice) The make the coefficients of the selected variable in both the equations equal by suitably multiplying or dividing one or both equations by suitable constants.

For ex., In the first set of equations, if you choose p, then multiply first equation by 3 and second equation by 10 so that coefficient of p becomes 30 in both. Such multiplication should be done THROUGHOUT the equation. Means, when first equation is multiplied bu=y 3, the other terms become 36q and 48 respectively.

Now suppose you are choosing the variable q, then multiply the second equation alone by 3 so that coefficient of q in both are 12.

It can be seen that the second choice leads to easier work-out. After some practice, you will naturally select the easier route.

Step 2.

Subtract one of the equation from the other. This eliminates the chosen variable as the coeff are same and we get an equation on one variable only. Determine the value of that remaining variable by transposition.

Step 3.

Substitute the value of the second variable calculated in step 2 in any one of the original equations or the third equation obtained after multiplication / division in step 1. After suitable transposition, the value of the first chosen variable will be obtained.

Please try to solve the above sets of equations on these lines. If you have still more difficulty, let us know, we will help further.

10p + 12q= 16

3p+ 4q=4

(3p+4q=4) x 3

=9p+12q=12

(10p+12q=16) – (9p+12q=12) = (p=4)

So p=4

Substitute into easiest equation:

(3x4) + 4q=4

12+4q=4

4-12= -8

4q= -8

Q= -2

P=4 and q =-2

Follow the pattern for the others.

I'll do one of them, the first one:

10p + 12q = 16

[3p + 4q = 4] • (-3) --> -9p - 12q = -12

p = 4

3(4) + 4q = 4

12 + 4q = 4

4q = -8

q = -2

(4, -2)

10p + 12q= 16

3p+ 4q=4

multiply second eq by -3 and add two eq together

10P+12q=16

-9p-12q=-12

--------------------

P =4 so plug in 4 for p: 3(4)+4q=4, q=-2

same principle for second set - multiply second eq by -2 and add

2s=-2, s=-1 and t=7

again, same for the last set. multiply the first by -5 and add

-7b=-70, b=10. a=-4

q in terms of p—2 equations:

10p + 12q = 16

5p + 6q = 8

q = (8 - 5p)/6

3p + 4q = 4

q = (4 - 3p)/4

Value of p:

4(8 - 5p) = 6(4 - 3p)

32 - 20p = 24 - 18p

2p = 8

p = 8/2 or 4

Value of q—1st equation:

= (8 - 5[4])/6

= (8 - 20)/6

= - 12/6 or - 2

Answer: p = 4, q = - 2

------------

t in terms of s—2 equations:

8s + 4t = 20

2s + t = 5

t = 5 - 2s

3s + 2t = 11

t = (11 - 3s)/2

Value of s:

2(5 - 2s) = 11 - 3s

10 - 4s = 11 - 3s

s = - 1

Value of t—1st equation:

= 5 - 2(- 1)

= 5 + 2

= 7

Answer: s = - 1, t = 7

------------

b in terms of a—2 equations:

3a + 2b = 8

b = (8 - 3a)/2

15a + 3b = - 30

5a + b = - 10

b = - 5a - 10

Value of a:

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

8 - 3a = - 10a - 20

7a = - 28

a = - 4

Value of b—1st equation:

= (8 - 3[- 4])/2

= (8 + 12)/2

= 20/2 or 10

Answer: a = - 4, b = 10