How do I work out Simultaneous equations?

Question

I've been trying for hours to work out simultaneous equations in my revision guide, but I still can't get the correct answers. On some occasions, why do you have to multiply and on others you have to add or subtract, this puzzles me. This topic is important, because it could crop up in my Intermediate terminal paper.

Can anyone please work out these two simultaneous equations, the method and the answer.

1) 4x + 7y = 10
2x + 3y = 3

2) 3a - 5b = 1
2a + 3b = 7

Any help or useful sites, will be greatly appreciated, thanks.

Answer 1

The whole trick in working out simultaneous equations is figuring out a way to eliminate all but one of the variables so that it can be solved for. Once you get one, you start working backward to get the other(s). While this operation can be accomplished in several ways, one of the most common is by elimination.

You can add or subtract equations without affecting the validity of the equality. You can also multiply or divide an entire equation by anything but zero without changing the equation. Elimination makes use of these properties to get rid of one or more variables.

In your first problem you can see that if we multiply the second equation by 2 and subtract it from the first equation, we will eliminate the x variable:

Original
4x + 7y = 10
2x + 3y = 3

Multiply second eq. by 2
4x + 7y = 10
4x + 6y = 6

Subtract second eq from first
1y = 4

Knowing that y = 4, go back to either equation and substitute 4 for y.

2x + 3y = 3
2x + 3(4) = 3
2x + 12 = 3
2x = -9
x = -4.5

To check, plug both numbers into the other equation
4(-4.5) + 7(4) = 10
-18 +28 = 10
Check!

Your second problem can be addressed in the same manner. In this case, we could multiply the second equation by 1.5 and subtract. An alternative approach would be to multiply the first equation by 2 and the second equation by 3 like this:

Original
3a - 5b = 1
2a + 3b = 7

Multiply first by 2 and second by 3
6a - 10b = 2
6a + 9b = 21

Subtract
-19b = -19
b = 1

Substitute back into the first
3a -5(1) = 1
3a = 6
a = 2

It doesn't matter which variable you eliminate first. In the first problem we could have multiplied the first eq. by 3 and the second eq. by 7 like this:

12x + 21y = 30
14x + 21y = 21

Subtract:
-2x = 9
x= -4.5

Same answer as before

Answer 2

Working Out Simultaneous Equations

Answer 3

Hey there,

Question 1:

4x + 7y = 10 -- Sentence 1
2x + 3y = 3 -- Sentence 2

You have two unknowns. Using Sentence 1, find x in terms of y.

4x + 7y = 10
4x = 10 - 7y
x = (10 - 7y)/4

Now that you have x, you can substitute it into Sentence 2 so that the equation only has one unknown, y.

[x = (10 - 7y)/4]

2x + 3y = 3
2 [(10 - 7y)/4] + 3y = 3

[(10 - 7y)/2] + 3y = 3 -- Multiply the whole equation by 2 (the denominator of the fraction) to eliminate the fraction
{[(10 - 7y)/2] + 3y = 3 } x 2

10 - 7y + 6y = 6 -- Now, group the like terms
-7y + 6y = 6 - 10 -- Solve for y

-y = -4
y = 4

Now that you've found y, you can substitute it into your first equation.

[y = 4]

x = (10 - 7y)/4
x = (10 - 7(4))/4
x = (10 - 28)/4
x = -18/4 -- Simplify
x = -9/2

**Check your answers using Sentence 2.

[x = -9/2; y = 4]

2x + 3y
= 2(-9/2) + 3(4)
= -9 + 12
= 3 (same as given information)

Therefore, when x = -9/2, y = 4

Question 2:

3a - 5b = 1 -- Sentence 1
2a + 3b = 7 -- Sentence 2

You have two unknowns. Using Sentence 1, find a in terms of b.

3a - 5b = 1
3a = 1 + 5b
a = (1 + 5b)/3

Now that you have a, you can substitute it into Sentence 2 so that the equation only has one unknown, b.

[a = (1 + 5b)/3]

2a + 3b = 7
2[(1 + 5b)/3] + 3b = 7

[(2 + 10b)/3] + 3b = 7 -- Multiply the whole equation by 3 (the denominator of the fraction) to eliminate the fraction
{ [(2 + 10b)/3] + 3b = 7 } x 3

2 + 10b + 9b = 21 -- Now, group the like terms
10b + 9b = 21 - 2 -- Solve for b

19b = 19
b = 19/19
b = 1

Now that you've found b, you can substitute it into your first equation.

[b = 1]

a = (1 + 5b)/3
a = (1 + 5(1))/3
a = 6/3
a = 2

**Check your answers using Sentence 2.

[a = 2; b = 1]

2a + 3b
= 2(2) + 3(1)
= 4 + 3
= 7 (same as information given)

Therefore, when a = 2, b = 11.

Hope this helps! :)

Answer 4

The idea of working out simultaneous equations its to eliminate on of the variables and then have one equation in one variable to solve. Once solved you can then put the amount of that variable in either of the 2 original equations and get the value of the other variable.

4 X + 7 Y = 10
2 X + 3 Y = 3

By multiplying the lower equation by -2 you get

-4 X - 6 Y = -6

When adding this to the top equation you get

0 X + 1 Y = 4 thus Y = 4

Now by entering Y = 4 in the first equation you get

4 X + 7(4) = 10 and
4 X + 28 = 10 and
4 X = 10 - 28 = - 18 so
X = -18/ 4 = -9/2

In second set You can see that by multiplying top equation by
2 and bottom equation by -3 you will get 2 equations in which if you add them will all0ow the (a) term to drop out leaving one equation with just a (b) term for which you can solve and then knowing the value of (b) you can substitute it into either equation and solve for (a)

Answer 5

#1)
4x + 7y = 10
2x + 3y = 3

We want to eliminate one variable, but in order to do that, their coefficients need to be opposites. I'll eliminate the 4x. The first equation is fine as written, but we'll need to multiply the second equation by (-2) to get the coefficients the same...

Equation 1: 4x + 7y = 10
Equation 2: -4x - 6y = -6

Now, add the two equation together and simplify...
4x - 4x + 7y - 6y = 10 - 6
y = 4

Now that you have that, put it into one of the equations to find x.

4x + 7y = 10
4x + 7(4) = 10
4x + 28 = 10
4x = -18
x = -9/2

================
#2
3a - 5b = 1
2a + 3b = 7

Since we have one positive and one negative coefficient on the "b", I'll eliminate that one. We need a common coefficient... 15 is the closest. So, multiply the first equation by 3 and the second equation by 5.

Equation 1: 9a - 15b = 3
Equation 2: 10a + 15b = 35

Now add them together and simplify:
9a + 10a - 15b + 15b = 3 + 35
19a = 38
a = 2

Now put that back into one of the equations to find b.

3a - 5b = 1
3(2) - 5b = 1
6 - 5b = 1
-5b = -5
b = 1

Answer 6

This Site Might Help You.

RE:
How do I work out Simultaneous equations?
I've been trying for hours to work out simultaneous equations in my revision guide, but I still can't get the correct answers. On some occasions, why do you have to multiply and on others you have to add or subtract, this puzzles me. This topic is important, because it could crop up in my...

Answer 7

equation2
need to solve on equation for either 'a' or 'b' then plug your answer into the other equation.2a+3b=7
3b=7-2a
b=1/3(7-2a)
b=7/3-2/3a
plug this into equation 1
3a-5b=1 where b=7/3-2/3a
3a-5(7/3-2/3a)=1
3a-35/3+10/3a-1=0
-38/3+19/3a=0
solve for a
a=2
plug this into either equation to get b
3a-5b=1 where a=2
3(2) -5b=1
solve for b
6-5b=1
-5b=-5
b=1

do the same way for equation1