# Solve the simultaneous equations: 2x + 3y = 28 and x + y = 11

As the number of equations is the same as the number of unknowns, there is exactly one solution!

We start by labelling the two equations:

2x + 3y = 28 (1)

x + y = 11 (2)

There is more than one way to approach this. We only need to use one approach, but let's consider two different methods here.

**Method 1**

We can use substitution. We start by making the coefficient of one of our unknown values the same in both equations; in the first equation, we have the term "2x" and in the second equation we have the term "x". We can multiply both sides of the second equation by 2 in order to have a term in "2x".

So let's start by multiplying the second equation by 2:

2*(2): 2x + 2y = 22

We can now subtract this from the first equation:

(1)-2*(2): 2x + 3y - 2x - 2y = 28 - 22

so y = 6

We've found y! To find x, we can substitute our value for y into either of the two equations. Let's substitute it into equation (2). We see:

x + 6 = 11

so x = 5

We have now found both unknowns. We know from the previous step that equation (2) is satisfied. In order to check our answer, it is a good idea to substitute both unknowns into equation (1):

LHS (the left hand side of the equation) = 2x + 3y

= 2*5 + 3*6

= 10 + 18

= 28

= RHS (the right hand side of the equation)

Both equations are satisfied, so we know that we have found the correct answer.

**Method 2**

We can solve simultaneous equations by elimination. We start by making one of our unknowns the subject of one of our equations. Let's make y the subject of equation (2). We simply subtract x from both sides, so:

y = 11 - x

We can now substitute this into equation (1); we write (11 - x) instead of y as they are the same. So (1) becomes:

2x + 3(11 - x) = 28

2x + 33 - 3x = 28

33 - x = 28

33 = 28 + x

x = 5

We've found x, and now we can find y and check our answer in the same way as in Method 1.