Question

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

Answer 1

Hint: The most straightforward way to solve any system of equations is the method of substitution. In short, the way to do it is to find the 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..

Complete Step by Step Solution:
Let’s apply this to our system. It looks like we can easily represent $y$ in terms of $x$ using the first equation
$ \Rightarrow 8{x^2} + 5y = 100$
Let, $y = \dfrac{{100 - 8{x^2}}}{5}$
Now substitute it into the second equation, hence we get
$\, \Rightarrow 6{x^2} - x - 3\dfrac{{100 - 8{x^2}}}{5} = 5$
The latter is a simple quadratic equation with one variable$x$that we, presumably, know how to solve.
First, we simplify it as follows:
$ \Rightarrow \dfrac{{54}}{5}{x^2} - x - 65 = 0$
Solutions to these equations are
$ \Rightarrow {x_{1,2}} = \dfrac{{1 \pm \sqrt {1 + 4 \times 65 \times \dfrac{{54}}{5}} }}{{2 \times \dfrac{{54}}{5}}}$
Now simplify this equation, we get
$ \Rightarrow {x_{1,2}} = (1 \pm 53) \times \dfrac{5}{{108}}$
First we add the equation and simplify the whole term, hence we get
$ \Rightarrow {x_1} = (1 + 53) \times \dfrac{5}{{108}} = 54 \times \dfrac{5}{{108}}$
$ \Rightarrow {x_1} = \dfrac{5}{2}$
Second we subtract the equation and simplify the whole term, hence we get
$ \Rightarrow {x_2} = (1 - 53) \times \dfrac{5}{{108}} = - 52 \times \dfrac{5}{{108}}$
$ \Rightarrow {x_2} = - \dfrac{{65}}{{27}}$
Now we can return to our representation of $y$ in terms of $x$ to find $y$ as follows
 \[ \Rightarrow {y_1} = \dfrac{{100 - 8x_1^2}}{5} = \dfrac{{100 - 8 \times {{\left( {\dfrac{5}{2}} \right)}^2}}}{5} = 10\]
 \[ \Rightarrow {y_2} = \dfrac{{100 - 8x_2^2}}{5} = \dfrac{{100 - 8 \times {{\left( { - \dfrac{{65}}{{27}}} \right)}^2}}}{5} = \dfrac{{7820}}{{729}}\]
So, we have two solutions to our system of equations
$ \Rightarrow ({x_1},{y_1}) = \left( {\dfrac{5}{2},10} \right)$
$ \Rightarrow ({x_2},{y_2}) = \left( { - \dfrac{{65}}{{27}},\dfrac{{7820}}{{729}}} \right)$

Note: Check the solutions:
Check$1$: Substituting$({x_1},{y_1})$into both equations and checking the identities.
$ \Rightarrow 8x_1^2 + 5{y_1} = 8 \times {\left( {\dfrac{5}{2}} \right)^2} + 5 \times 10$
Square the term of$\dfrac{5}{2}$and multiply by square the term by$8$and multiplies the second term. After we add the two-terms.
$ \Rightarrow 8 \times \dfrac{{25}}{4} + 5 \times 10 = 50 + 50 = 100$
$ \Rightarrow 6x_1^2 - {x_1} - 3{y_1} = 6 \times {\left( {\dfrac{5}{2}} \right)^2} - \dfrac{5}{2} - 3 \times 10$
Square the term of$\dfrac{5}{2}$and multiply the square term by $6$ and multiply the third term$3$by$10$. After we subtract the two numbers.
$ \Rightarrow \dfrac{{75}}{2} - \dfrac{5}{2} - 3 \times 10 = 35 - 30 = 5$
Both equations check.
Check:
Substituting$({x_2},{y_2})$ into both equations and checking the identities
$ \Rightarrow 8x_2^2 + 5{y_2} = 8 \times {\left( { - \dfrac{{65}}{{27}}} \right)^2} + 5 \times \dfrac{{7820}}{{729}}$
Square the term of \[ - \dfrac{{65}}{{27}}\] and take the common term in the denominator. We add the numerator and divide by the denominator
$ \Rightarrow \dfrac{{8 \times {{65}^2} + 5 \times 7820}}{{729}} = \dfrac{{33800 + 39100}}{{729}}$
$ \Rightarrow \dfrac{{72900}}{{729}} = 100$
$ \Rightarrow 6x_2^2 - {x_2} - 3 \times {y_2} = 6 \times {\left( { - \dfrac{{65}}{{27}}} \right)^2} - \left( { - \dfrac{{65}}{{27}}} \right) - 3 \times \left( {\dfrac{{7820}}{{729}}} \right)$
Square the term of \[ - \dfrac{{65}}{{27}}\] and take the common term in the denominator. We add and subtract the numerator and divide by the denominator
$ \Rightarrow \dfrac{{25350}}{{729}} + \dfrac{{65}}{{27}} - \dfrac{{23460}}{{729}} = \dfrac{{25350 + 1755 - 23460}}{{729}}$
$ \Rightarrow \dfrac{{3645}}{{729}} = 5$
Both equations check.