Solve System of Equations: Finding (a+by)²(a-by) Using Completing the Square

System of Equations with Substitution Technique

{a2x3+2abx2y=2a(ax+by)a(x+1)=by \begin{cases} a^2x^3+2abx^2y=-2a(ax+by)\\ a(x+1)=-by \end{cases} b0,a0 b\neq0 ,\hspace{4pt}a\neq0

Solve the above equations by completing the square/ factoring:

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


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Step-by-step written solution

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1

Understand the problem

{a2x3+2abx2y=2a(ax+by)a(x+1)=by \begin{cases} a^2x^3+2abx^2y=-2a(ax+by)\\ a(x+1)=-by \end{cases} b0,a0 b\neq0 ,\hspace{4pt}a\neq0

Solve the above equations by completing the square/ factoring:

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


2

Step-by-step solution

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

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

This is without solving the system of equations,

Let's examine the given system:


{a2x3+2abx2y=2a(ax+by)a(x+1)=by \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 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)=byax+a=byax+by=a ( 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):

{a2x3+2abx2y=2a(ax+by)a(x+1)=byax+by=aa2x3+2abx2y=2a(a)a2x3+2abx2y=2a2 \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:

a2x3+2abx2y=2a2x(a2x2+2abxy)=2a2 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:

{x(a2x2+2abxy)=2a2ax+by=a \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(a2x2+2abxy)=2a2x((ax)2+2axby)=2a2x((ax)2+2axby+(by)2(by)2)=2a2x((ax)2+2axby+(by)2b2y2)=2a2x((ax+by)2b2y2)=2a2 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:

{x(a2x2+2abxy)=2a2ax+by=a{x((ax+by)2b2y2)=2a2ax+by=ax((a)2b2y2)=2a2x(a2b2y2)=2a2 \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(aby)=? (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):

{x(a2b2y2)=2a2ax+by=a \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:

{x(a2b2y2)=2a2ax+by=a{x(a2(by)2)=2a2ax=(a+by)/:a{x(a+by)(aby)=2a2x=a+bya \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:

{x(a+by)(aby)=2a2x=a+byaa+bya(a+by)(aby)=2a2(a+by)(a+by)(aby)a=2a2/(a)(a+by)2(aby)=2a3 \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

3

Final Answer

2a3 -2a^3

Key Points to Remember

Essential concepts to master this topic
  • Key Insight: Express one variable combination from the second equation
  • Substitution: Replace ax+by ax+by with a -a in first equation
  • Verification: Check by substituting final answer back into original expression ✓

Common Mistakes

Avoid these frequent errors
  • Attempting to solve for individual variables x and y
    Don't try to find x and y separately = unnecessary complexity and potential errors! The problem asks for a specific expression value, not individual variables. Always look for patterns and substitution opportunities to avoid solving the full system.

Practice Quiz

Test your knowledge with interactive questions

\( ax-3=1 \)

Without solving the equation, calculate the value of the following expression:

\( a^2x^2-6ax+14 \)

FAQ

Everything you need to know about this question

Why don't we need to solve for x and y individually?

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The problem asks for the value of (a+by)2(aby) (a+by)^2(a-by) , not the individual variables. By using clever substitution from the second equation, we can find this expression's value directly!

How did we know to use completing the square?

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After substitution, we had a2x2+2abxy a^2x^2 + 2abxy . This looks like part of (ax+by)2 (ax + by)^2 , so we completed the square by adding and subtracting b2y2 b^2y^2 .

What if I get confused with all the algebra steps?

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Break it down! First find what ax+by ax+by equals from equation 2. Then substitute this value everywhere you see ax+by ax+by . Take each step slowly and double-check your algebra.

How do I recognize when to use difference of squares?

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Look for expressions like a2b2y2 a^2 - b^2y^2 . This factors as (a+by)(aby) (a+by)(a-by) . When you see a square minus square, think difference of squares!

Why is the final answer negative?

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From the second equation, we found x=a+bya x = -\frac{a+by}{a} . When we substitute this into our factored form, the negative sign carries through to give us 2a3 -2a^3 .

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