Problem 51 - Precalculus with Limits

Question

Choosing a Solution Method In Exercises \(49-56\) , solve the system graphically or algebraically. Explain your choice of method. $$\left\\{\begin{array}{l}{x-2 y=4} \\ {x^{2}-y=0}\end{array}\right.$$

Step-by-Step Solution

Answer

The solutions to the system of equations are (-4,-4) and (6,1).

Step 1: Rearrange the first equation

Rearrange the first equation to solve for x. Which is done as follows: \(x = 4+2y\)

Step 2: Substitute x in the second equation

Now, take the solution for x from the first equation and substitute it into the second equation, getting: \((4+2y)^2 -y = 0\)

Step 3: Simplify the equation

When the second equation is simplified, it now becomes: \(4y^2 +16y +16 - y = 0\). Further simplify to: \(4y^2 + 15y + 16 = 0\)

Step 4: Solve for y

Here, you solve for \(y\) by factoring the equation, resulting in: \((4y+16)(y-1)=0\). Therefore, \(y=-4\) or \(y=1\) are the solutions for \(y\).

Step 5: Solve for x

Finally, substitute the \(y\) values into the rearranged first equation \(x = 4+2y\) to find the values of \(x\). When \(y=-4\), \(x=4-8=-4\). When \(y=1\), \(x=4+2*1=6\). Therefore, the solutions in the form of coordinate pairs are \((-4,-4)\) and \((6,1)\).

Key Concepts

graphical solutions

Graphical solutions involve visualizing systems of equations on a set of axes to determine where their graphs intersect.
By plotting each equation in the system on the same coordinate plane, the point(s) of intersection represent the solution(s) to the system.

• The first equation, \(x - 2y = 4\), can be rearranged to slope-intercept form as \(y = \frac{x}{2} - 2\). This represents a straight line.
• The second equation, \(x^2 - y = 0\), can be rearranged to \(y = x^2\), which is a parabola opening upwards.

When graphed, a straight line and a parabola might intersect at one or more points, depending on their relative positions.
In this case, the graphical method might not always be the most precise due to potential scaling issues or visual inaccuracies, especially when dealing with non-linear functions like the parabola in this example. However, it quickly offers a visual intuition of potential solutions.

algebraic solutions

Algebraic solutions use algebraic manipulations to find the exact solution(s) to a system of equations. This method involves techniques such as substitution or elimination to simplify and solve the equations.
One benefit of algebraic solutions is precision, offering exact numbers as results, unlike the graphical method that might be prone to visual misinterpretation.

• In our system, we can start by solving one of the equations for a single variable. For instance, solve the first equation for \(x\), getting \(x = 4 + 2y\).
• Substitute this expression for \(x\) into the second equation \(x^2 - y = 0\).
• This substitution gives a single equation in terms of \(y\): \((4 + 2y)^2 - y = 0\).

After simplifying, you can solve this equation to find the values of \(y\). Once \(y\) is known, substitute back into the expression for \(x\) to find the corresponding values of \(x\). Algebraic solutions provide clear paths to understanding the interplay of equations, particularly when exact calculation is necessary.

factoring polynomials

Factoring polynomials is a critical algebraic technique used to simplify and solve polynomial equations.
In this problem, once the substitution and simplification lead to a polynomial equation in terms of \(y\), factoring becomes essential.

• Starting with the equation \(4y^2 + 15y + 16 = 0\), you need to factor it to find the roots.
• Factoring may involve finding numbers that both add to give the middle term (in this case, 15y) and multiply to the constant term (in this case, 64, because \(4 \times 16\)).

After factoring, \((4y + 16)(y - 1) = 0\), set each factor to zero: \(4y + 16 = 0\) or \(y - 1 = 0\).

• Solving \(4y + 16 = 0\) gives \(y = -4\).
• Solving \(y - 1 = 0\) gives \(y = 1\).

Each of these values of \(y\) can be used to find the respective \(x\) coordinates in the system's solution. Factoring provides clear insights and a direct path to finding roots, essential for solving quadratic equations.