1. ## Simultaneous linear equations

Hi guys I am currently having difficulties with these simultaneous equations, where you have to find out what x is and what y is, i am really confused with what methods you are meant to use for each one. PLEASE HELP I WOULD REALLY APPRECIATE IT THANKS!

1)
4x+3y=3
6x+5y=7

2)
5x-3y=7
3x+5y=-6

3) x+y=4
y=x+2

2. So there are two really good techniques that I know of for solving systems of linear equations. They are substitution and subtraction (this may have a different name). There are more advanced techniques for solving large systems but that can be left for another time and place. Anyways either will work for any of the problems below but one may be easier and require fewer steps than the other depending on the problem.

For problem 1 let's use substitution:

$\displaystyle 4x+3y=3$

$\displaystyle 6x+5y=7$

Take either one of your equations and solve for one variable in terms of the other (doesn't matter which equation or which variable you choose!). I will choose the first equation and solve for $\displaystyle x$ in terms of $\displaystyle y$.

$\displaystyle 4x+3y = 3$

$\displaystyle 4x = 3-3y$

$\displaystyle x = \frac{3-3y}{4}$

Now substitute x into the other equation:

$\displaystyle x = \frac{3-3y}{4}$

$\displaystyle 6x + 5y = 7$

$\displaystyle 6(\frac{3-3y}{4})+5y = 7$

$\displaystyle \frac{18-18y}{4} + 5y = 7$

$\displaystyle \frac{18}{4}-\frac{18y}{4} + 5y = 7$

$\displaystyle \frac{y}{2} = \frac{5}{2}$

$\displaystyle y=5$

Now that you have found $\displaystyle y$, you can use the equation for $\displaystyle x$ to find its value:

$\displaystyle y=5$

$\displaystyle x = \frac{3-3y}{4}$

$\displaystyle x = \frac{3-15}{4}$

$\displaystyle x = \frac{-12}{4}$

$\displaystyle x = -3$

We can also use the technique of subtraction. The idea here is to get the coefficient in front of one of the variables to be the same in both equations. Then you can subtract the equations one from the other, thus eliminating one variables.

For problem 2:

$\displaystyle 5x-3y=7$

$\displaystyle 3x+5y=-6$

Multiply the first equation by 3,and the second equation by 5. Then

$\displaystyle 15x-9y=21$

$\displaystyle 15x+25y = -30$

Notice that the coefficient on the front of the x variable is the same in both equations. Now subtract the x's, subtract the y's and subtract the numbers on the right hand side of the equation. We get:

$\displaystyle 15x-9y=21$

$\displaystyle \underline{-(15x+25y = -30)}$

$\displaystyle 0x - 34y = 51$

Now we can find y, substitute it back into one of our original equations and find x.