Solve the simultaneous equations, (1) 4x+y=23 and (2) 3x+5y=111/2

Simultaneous equations are equations that have one or more unknown values. There are a multitude of ways to solve simultaneous equations, however the elimination method is the most common at GCSE level maths. Currently neither of the variables/unknowns are equal and so the elimination method cannot be used yet. However, you can do this by equating either the x or y values. In this case however, i will multiply (1) by 5 in order for both the equations to have 5y and then the elimination method can be used. (1) will now equal 20x+5y=115. While (2) remains the same.It can now be seen that both equations have the same coefficient for their unknown value y. Now we can minus equation (2) from (1) to eliminate the y value, and leave us with the x value which can then be solved. (20x+5y=115)-(3x+5y=111/2).This leaves us with 17x=119/2. If you then divide through by 17 then the x-value has been solved. x=3.5.However, we need to find out the y value, which can be found through substituting the x value (3.5) back into either of the original questions. This can be done by 4(3.5)+y=23. Which equals 14+y=23. Now subtract 14 to find out that y=9.Therefore x=3.5, y=9.