Question

How do you find the solution of the system of equations `8x^2 + 5y =100` and `6x^2 -x -3y =5`?

Answer 1

The most straight forward (but not always the best) way to solve any system of equations is the method of substitution. In short, the way to do it is to find a simplest equation, from which you can represent a value of one of the variables in terms of all others, and substitute this into all other equations, thus reducing the number of equations and the number of variables. Then do it again and again until you are left with only one equation with one variable.

Let's apply this to our system. It looks like we can easily represent `y` in terms of `x` using the first equation `8x^2+5y=100`:
`y=(100-8x^2)/5`

Now substitute it into the second equation getting
`6x^2-x-3(100-8x^2)/5=5`

The latter is a simple quadratic equation with one variable `x` that we, presumably, know how to solve. First, we simplify it as follows:
`54/5 x^2 - x -65 = 0`

Solutions to this equations are
`x_(1,2)=[1+-sqrt(1+4*65*54/5)]/(2*54/5)=(1+-53)*5/108`
or
`x_1=5/2`
`x_2=-65/27`

Now we can return to our representation of `y` in terms of `x` to find `y` as follows:
`y_1=(100-8x_1^2)/5=[100-8*(5/2)^2]/5=10`
`y_2=(100-8x_2^2)/5=[100-8*(-65/27)^2]/5=7820/729`

So, we have two solutions to our system of equations:
`(x_1,y_1)=(5/2,10)`
`(x_2,y_2)=(-65/27, 7820/729)`

Always check the solutions.

Check 1. Substituting `(x_1,y_1)` into both equations and checking the identities.
`8x_1^2+5y_1=8*(5/2)^2+5*10=`
`=8*25/4+5*10=50+50=100`

`6x_1^2-x_1-3*y_1=6*(5/2)^2-5/2-3*10=`
`=75/2-5/2-3*10=35-30=5`
Both equations check.

Check 2. Substituting `(x_2,y_2)` into both equations and checking the identities.
`8x_2^2+5y_2=`
`=8*(-65/27)^2+5*7820/729=`
`=(8*65^2+5*7820)/729=`
`=(33800+39100)/729=`
`=72900/729=100`

`6x_2^2-x_2-3*y_2=`
`=6*(-65/27)^2-(-65/27)-3*7820/729=`
`=25350/729+65/27-23460/729=`
`=(25350+65*27-23460)/729=`
`=(25350+1755-23460)/729=`
`=3645/729=5`
Both equations check.