Systems of Quadratic Equations - Examples, Exercises and Solutions

First, recall the formula for calculating the area of a rectangle:

The area of a rectangle (which has two pairs of equal opposite sides and all angles are $90\degree$) with sides of length $a,\hspace{4pt} b$ units, is given by the formula:

$\boxed{ S_{\textcolor{blue}{\boxed{\hspace{6pt}}}}=a\cdot b }$(square units)

$$

After recalling the formula for the area of a rectangle, let's proceed to solve the problem:

Begin by denoting the area of the given rectangle as: $S_{\textcolor{blue}{\boxed{\hspace{6pt}}}}$ and proceed to write (in mathematical notation) the given information:

$S_{\textcolor{blue}{\boxed{\hspace{6pt}}}}=x^2-13$

Continue to calculate the area of the rectangle given in the problem:

$$

Using the rectangle area formula mentioned earlier:

$S_{\textcolor{blue}{\boxed{\hspace{6pt}}}}=a\cdot b\\ \downarrow\\ S_{\textcolor{blue}{\boxed{\hspace{6pt}}}}=(x-4)(x+1)$

Continue to simplify the expression that we obtained for the rectangle's area, using the distributive property:

$(c+d)(h+g)=ch+cg+dh+dg$

We are able to obtain the area of the rectangle by

using the distributive property as shown below:

$S_{\textcolor{blue}{\boxed{\hspace{6pt}}}}=(x-4)(x+1) \\ S_{\textcolor{blue}{\boxed{\hspace{6pt}}}}=x^2+x-4x-4\\ \boxed{ S_{\textcolor{blue}{\boxed{\hspace{6pt}}}}=x^2-3x-4}$

Recall the given information:

$S_{\textcolor{blue}{\boxed{\hspace{6pt}}}}=x^2-13$

Therefore, we can conclude that:

$x^2-3x-4=x^2-13 \\ \downarrow\\ -3x=4-13\\ -3x=-9\hspace{6pt}\text{/}(-3)\\ \boxed{x=3}$

We solved the resulting equation simply by combining like terms, isolating the expression with the unknown on one side and dividing both sides by the unknown's coefficient in the final step,

Note that this result satisfies the domain of definition for x, which was given as:

-1\text{<}x\text{<}4 and therefore it is the correct result

The correct answer is answer C.

In this problem we are required to calculate the value of the expression:

$(a+by)^2(a-by)=\text{?}$

This is without solving the system of equations,

Let's examine the given system:

$\begin{cases} a^2x^3+2abx^2y=-2a(ax+by)\\ a(x+1)=-by \end{cases}$

In this system there are two equations with two unknowns and two parameters, one of the equations is of a higher degree, and the second is of first degree,

From which we need to calculate the value of the requested expression, without solving the system,

Let's try to find hints in the question and find another "trick", for this we can first notice that the expression:

$ax+by$

Which appears inside the multiplication on the right side in the first equation, can also be found in the second equation, by opening parentheses and moving terms:

$( a(x+1)=-by\\ \downarrow\\ ax+a=-by\\ \boxed{ax+by=-a}$

In other words, we can substitute the value of this expression completely in the first equation, let's do this (for convenience in calculation we'll show the complete system of equations):

$\begin{cases} a^2x^3+2abx^2y=-2a\textcolor{red}{(ax+by)}\\ a(x+1)=-by\rightarrow \boxed{\textcolor{red}{ax+by=-a}} \end{cases} \\ \downarrow\\ a^2x^3+2abx^2y=-2a\textcolor{red}{(-a)}\\ \boxed{a^2x^3+2abx^2y=2a^2}$

We have therefore received a simpler equation, but still of a higher degree with two unknowns, let's try to reduce the degree of the expression on the left side,

Note that in this expression we can factor out a common term:

$a^2x^3+2abx^2y=2a^2\\ \downarrow\\ x(a^2x^2+2abxy)=2a^2$

Now let's look at the expression in parentheses, and think again about the second equation we got in the system of equations:

$\begin{cases} x(\textcolor{blue}{a^2x^2+2abxy})=2a^2\\ ax+by=-a \end{cases} \\$

We can see that, by completing the square in the expression, we can use the second equation again, this is done by presenting the expression in parentheses as a binomial squared plus a correction term, remember that for this we need to identify the missing term in the expression so we can complete the square, and then add and subtract it from the expression:

$x(a^2x^2+2abxy)=2a^2\\ \downarrow\\ x\big((ax)^2+2axby\big)=2a^2\\\\ x\big((ax)^2+2axby+\underline{\underline{(by)^2-(by)^2}}\big)=2a^2\\ x\big((\textcolor{red}{ax})^2+2\textcolor{red}{ax}\textcolor{green}{by}+(\textcolor{green}{by})^2-b^2y^2\big)=2a^2\\ \downarrow\\ x\big((\textcolor{red}{ax}+\textcolor{green}{by})^2-b^2y^2)=2a^2$

Let's now show the system of equations we got after the last step, then we'll substitute the value of the expression from the second equation into the first equation (bottom line) and simplify:

$\begin{cases} x(a^2x^2+2abxy)=2a^2\\ ax+by=-a \end{cases} \\ \downarrow\\ \begin{cases} x\big((\underline{ax+by})^2-b^2y^2\big)=2a^2\\ ax+by=-a \end{cases} \\ \downarrow\\ x\big((\underline{-a})^2-b^2y^2\big)=2a^2\\ x(a^2-b^2y^2)=2a^2$

Now we can examine our goal - calculating the value of the expression:

$(a+by)^2(a-by)=\text{?}$

We can see that we are already very close to getting the value of this expression from the equivalent system of equations we received (we'll also note the second equation):

$\begin{cases} x(a^2-b^2y^2)=2a^2\\ ax+by=-a \end{cases} \\$

We can now use factoring using the difference of squares formula, and factor the expression in parentheses, we can also now isolate x from the second equation, let's do this in parallel:

$\begin{cases} x(a^2-b^2y^2)=2a^2\\ ax+by=-a \end{cases} \\ \downarrow\\ \begin{cases} x(a^2-(by)^2)=2a^2\\ ax=-(a+by)\hspace{6pt}\text{/}:a \end{cases} \\ \downarrow\\ \begin{cases} x(a+by)(a-by)=2a^2\\ x=-\frac{a+by}{a} \end{cases} \\$

Now let's substitute the isolated unknown in the first equation, identify the requested expression, and calculate its value by isolating it:

$\begin{cases} \textcolor{blue}{x}(a+by)(a-by)=2a^2\\ \textcolor{blue}{x=-\frac{a+by}{a}} \end{cases} \\ \downarrow\\ -\frac{a+by}{a}\cdot(a+by)(a-by)=2a^2 \\ -\frac{(a+by)(a+by)(a-by)}{a}=2a^2\hspace{6pt}\text{/}\cdot (-a)\\ \boxed{(a+by)^2(a-by)=-2a^3}$

Therefore the correct answer is answer D