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 1We 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 = 22We can now subtract this from the first equation:(1)-2*(2): 2x + 3y - 2x - 2y = 28 - 22so y = 6We'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 = 11so x = 5We 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= 25 + 36= 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 2We 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 - xWe 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) = 282x + 33 - 3x = 2833 - x = 2833 = 28 + xx = 5We've found x, and now we can find y and check our answer in the same way as in Method 1.

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