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How can i solve the following simultaneous equations? 5x + y = 4 and 3x + 2y = 5?

There are two ways to approach this problem. In face there are two ways to approach every single set of simultanious equation. Firstly we can try eliminating terms. Or we can also try substituting terms. Lets take a look at both now.

Starting off with the elimination method. We seek to get the same number of a single unknown. May it be x or y. So in this example, lets try to get 2y in both equations as its the easier than obtaining the same number of x's. So we would obtain the following;

10x + 2y = 8 (we multiplied every term in the first equation by 2)

3x + 2y = 5

We can now subtract (3x + 2y = 5) from (10x + 2y = 8) to result in (7x + 0y = 3). This is done on a term by term basis. 10x - 3x, 2y - 2y and 8 - 5 would be the right calculations to do. Now we can see that 7x = 3. So x = 3/7. We can now use our algebra skills to substitute the known x into one of the equations earlier.

As we know that x = 3/7, we can now say 3(3/7) + 2y = 5. By using algebra, we can rearrange to obtain the following;

9/7 + 2y = 5 (Multiply 3 by 3/7)

2y = 16/5 (Subtract 9/7 from 5)

y = 8/5

Now lets take a look at the substitution method. For this method, we need to isolate one of the unknowns so that there is only of it. So for example rewrite 5x + y = 4 in such a way that we show x = something. Lets find it.

5x + y = 4

x + y/5 = 4/5 (divide every term by 5)

x = 4/5 - y/5 (take y/5 to the other side of the = sign)

Now we would need to substitute this expression into the other simultanious equation like so.

3x + 2y = 5 becomes

3(4/5 - y/5) + 2y = 5

12/5 - 3y/5 + 2y = 5

By futher simplifying and rearranging, we obtain that y = 8/5. Therefore we can now substitute the known y into one of the equatoins above to obtain x. Just like in the previous method. So substituting y = 8/5 into 3x + 2y = 5 would lead us to finding out that x = 3/7

That's how we find the solutions to simultanious equations.

Tarun S. GCSE Maths tutor, A Level Maths tutor, Mentoring HTML & CSS ...

1 year ago

Answered by Tarun, a GCSE Maths tutor with MyTutor


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