Problem 5 - College Algebra Essentials

Question

In Exercises \(1-18,\) solve each system by the substitution method. $$ \left\\{\begin{array}{l} y=x^{2}-4 x-10 \\ y=-x^{2}-2 x+14 \end{array}\right. $$

Step-by-Step Solution

Answer

The solutions to the given system of equations are \((4,-6)\) and \((-3,5)\)

Step1: Express one variable in terms of the other variable

In this case, both equations are already solved for \(y\). Therefore, you can set the two expressions equal to each other to eliminate \(y\). This gives \(x^{2}-4x-10=-x^{2}-2x+14\)

Step 2: Simplify the equation

Collect like terms on one side of the equation, resulting in \(2x^{2}-2x-24=0\). This can be further simplified by dividing through by 2, resulting in: \(x^{2}-x-12=0\)

Step 3: Solve the quadratic equation

Factor the quadratic equation: \((x-4)(x+3)=0\). Setting each factor equal to zero gives the solutions \(x=4\) and \(x=-3\)

Step 4: Find corresponding \(y\) values

Substituting \(x=4\) into the first equation (you could also use the second one, it wouldn’t make a difference), we get \(y=4^{2}-4*4-10=-6\). Similarly, substituting \(x=-3\), we get \(y=(-3)^{2}+3*4-10=5\)

Step 5: Write the solution as ordered pairs

The solutions to the system of equations can be written as ordered pairs \((x,y)\). So, the solutions to the system of equations are \((4,-6)\) and \((-3,5)\)

Key Concepts

System of Equations

A system of equations consists of multiple equations that share a set of variables. The goal is to find values for these variables that satisfy all the equations simultaneously. In our example, we have two equations involving the variables x and y:

• \(y = x^2 - 4x - 10\)
• \(y = -x^2 - 2x + 14\)

Both equations express y in terms of x. Therefore, by using the substitution method, we equate the right-hand sides, allowing us to eliminate y and solve for x.

Quadratic Equations

Quadratic equations have the form \(ax^2 + bx + c = 0\). Here, an equation contains a variable raised to the second power, representing the main characteristic of quadratics. In our example, after setting the expressions equal to remove the y variable, a quadratic equation is formed: \[x^2 - x - 12 = 0\]This matches the standard form of a quadratic equation with \(a=1\), \(b=-1\), and \(c=-12\). Solving quadratic equations involves finding values of x that make the equation true.

Factoring

Factoring is a crucial step in solving quadratic equations by breaking down the expression into simpler terms that can be multiplied together to obtain the original quadratic. For instance, consider the equation:\[x^2 - x - 12 = 0\]Identifying the factors involves finding two numbers whose product is \(-12\) and sum is \(-1\). These numbers are \(4\) and \(-3\). Using these, we factor the quadratic as: \[(x - 4)(x + 3) = 0\]Each factor represents a potential solution for x.

Solving Equations

Once the quadratic equation is factored, we solve for x by setting each factor to zero:

• \(x - 4 = 0\) leads to \(x = 4\)
• \(x + 3 = 0\) leads to \(x = -3\)

With the x values known, we can substitute back into either of the original equations to find corresponding y values. This results in the ordered pairs

• \((4, -6)\)
• \((-3, 5)\)

These pairs are the solutions to our original system, representing where both equations hold true simultaneously.