Figuring $x$ and $y$ from two linear equations

Question

I have a mini exam in a month to study for and I'm looking at systems of equations at the moment. I have this question to look at right now:

Find $x$ and $y:$

$x-5y+4=1$

$\dfrac{x+1}{2}=y^2$

Now normally I would make a substitution into one of the equations (probably solve the first for y and then insert this into the second.)

However I'm not sure how do things straight off with a $y^2$ component.

I would probably multiply the left hand side by two to give $x+1=2y^2$

Could anybody kindly offer me any wisdom on figuring, considering I have the squared part?

Answer 1

Solve $x$ in your first equation:

• $$x-5y+4=1\Longleftrightarrow x=1-4+5y=5y-3$$

Substitute it in, in the second one and solve $y$:

• $$\frac{5y-3+1}{2}=y^2\Longleftrightarrow$$ $$\frac{5y-2}{2}=y^2\Longleftrightarrow$$ $$5y-2=2y^2\Longleftrightarrow$$ $$-2y^2+5y-2=0$$

Using the quadratic formula we can find that:

$$y=\frac{5\pm\sqrt{9}}{4}=\frac{5\pm3}{4}$$

So, now we can solve for $x$:

$$x=5\cdot\frac{5\pm3}{4}-3=\frac{25\pm15}{4}-3$$

Now, the solutions are:

• $$x_1=7,y_1=2$$
• $$x_2=-\frac{1}{2},y_2=\frac{1}{2}$$

Answer 2

From second equation

$$x+1=2y^2\implies x=2y^2-1$$

and now substitute en equation one:

$$(2y^2-1)-5y+3=0\iff 2y^2-5y+2=0$$

Solve this easy quadratic and get the possible value(s) of $\;y\;$ and then substitute above for the values of $\;x\;$