Problem 14 - Algebra and Trigonometry

Question

Solve each system by the substitution method. \(\left\\{\begin{array}{c}{2 x+5 y=1} \\ {-x+6 y=8}\end{array}\right.\)

Step-by-Step Solution

Answer

The solutions to the system are (2,2) and (-2,-2).

Step 1: Isolate a variable

Begin by isolating one of the variables in one of the equations. The first equation \(xy=4\) can be rewritten as \(y=\frac{4}{x}\) by dividing both sides by x.

Step 2: Substitute

Substitute \(y=\frac{4}{x}\) into the second equation \(x^2+y^2=8\). This leads to \(x^2+(\frac{4}{x})^2=8\), which simplifies to \(x^4-8x^2+16=0\) after multiplying all terms by \(x^2\) to clear the fraction and bringing all terms to the left.

Step 3: Solve the equation

The equation \(x^4-8x^2+16=0\) is a quadratic in terms of \(x^2\). We let \(u=x^2\) and end up with \(u^2-8u+16=0\). Solving for u gives \(u=4\) since this is a perfect square trinomial. Substituting back \(u=x^2\) gives \(x^2=4\), thus \(x=\pm 2\).

Step 4: Find y

Substitute \(x=\pm 2\) into the equation \(y=\frac{4}{x}\) to find the corresponding y-values. This gives \(y=\pm 2\). Therefore, the solutions to the system are (2,2) and (-2,-2).

Key Concepts

Systems of Equations

A system of equations involves finding values that satisfy multiple equations simultaneously. In this exercise, we deal with two equations:

• First equation: \(xy=4\)
• Second equation: \(x^2+y^2=8\)

To solve them, we aim to find an \((x,y)\) pair that satisfies both equations. An effective way to approach this is using the **substitution method**, where we solve one equation for one variable and substitute it into the other equation.
This method simplifies the problem, allowing you to focus on one variable at a time. The key advantage is reducing a system of equations to a single equation in one variable, making it easier to solve.
By substituting \(y=\frac{4}{x}\) into the second equation, we reduce the need to deal with two variables at once and focus on \(x\) first, a crucial step in such problems.

Quadratic Equations

A quadratic equation is a type of polynomial equation where the highest exponent of the variable is 2. In this exercise, after substitution, we end up with the equation \(x^4-8x^2+16=0\).
We noticed that this equation is quadratic in terms of \(x^2\), creatively using a substitution: Let \(u = x^2\). This turns our original equation into \(u^2-8u+16=0\).
Solving quadratic equations often involves methods such as factoring, using the quadratic formula, or recognizing special patterns like perfect squares. Here, recognizing a pattern can significantly simplify the solving process.

Perfect Square Trinomial

A perfect square trinomial is a special form of quadratic. It takes the form \(a^2 - 2ab + b^2 = (a-b)^2\). Recognizing these can make solving problems easier.

• In our exercise: \(u^2 - 8u + 16 = (u - 4)^2\)

Recognizing this as a perfect square allows us to factor directly, leading to solutions quickly without needing the quadratic formula.
Here, we see \((u-4)^2=0\), solving this gives \(u=4\). This step simplifies the process as perfect squares factor easily and predictably.