Problem 7 - College Algebra

Question

Solve each system by the substitution method. \(\left\\{\begin{array}{l}x+3 y=8 \\ y=2 x-9\end{array}\right.\)

Step-by-Step Solution

Answer

The solutions for the system of equations are \((-3, -4)\) and \((4, 3)\)

Step 1: Solving for x in the Second Equation

In the equation \(x - y = 1\), x can be isolated by adding y to both sides. The result is \(x = y + 1\)

Step 2: Substitute x in the First Equation

Substitute \(y + 1\) for x in the first equation, which is \(x^2 + y^2 = 25\). This gives \((y+1)^2 + y^2 = 25\)

Step 3: Expand and Simplify the Equation

Expanding and simplifying, we get \(y^2 + 2y + 1 + y^2 = 25\), then combining like terms we get \(2y^2 + 2y - 24 = 0\)

Step 4: Solve for y

Dividing the entire equation by 2 gives \(y^2+y-12=0\). Here, we can solve for y by factoring, we have \((y+4)(y-3)=0\). Setting each factor equal to zero gets the possible solutions \(y = -4\) and \(y = 3\)

Step 5: Solve for x

Finally, substitute each y-value into the original second equation to get x. Notice that \(x = y + 1\). Substituting \(y = -4\) gives \(x = -4 + 1 = -3\) and for \(y = 3\), we have \(x = 3 + 1 = 4\)

Step 6: Solutions for the System of Equations

Therefore, the solutions for the system of equations are \((-3, -4)\) and \((4, 3)\)

Key Concepts

Understanding Systems of Equations

When you have a **system of equations**, it means you're dealing with more than one equation at a time. The key goal is to find values for the variables that make all the equations true simultaneously. A solution to the system does just that—it’s a set of values that satisfies every equation in the group.

In our case, the system is made up of two equations:

• A quadratic equation: \(x^2 + y^2 = 25\)
• A linear equation: \(x - y = 1\)

The solution to this system is a pair of values (x, y) that simultaneously satisfy both equations. Solving systems of equations can be done by several methods — substitution being one of the simplest when one equation is easily rearranged.

Quadratic Equation Basics

Quadratic equations are equations of the second degree, meaning they include terms up to \(x^2\). In our system, the quadratic part comes in the form of \(x^2 + y^2 = 25\). Solving quadratic equations is a core part of many algebra problems.

These equations can have:

• Real solutions when the curve intersects the x-axis.
• Complex solutions when it doesn't intersect the x-axis.

In our exercise, the quadratic equation gives us a circle with a radius of 5 centered at the origin (0,0). The solutions of this circle with other lines or curves can represent real-world scenarios, such as points of intersection or maximum/minimum values.

The Steps to Finding Solutions of Equations

Finding the **solution of equations** in a system involves identifying values that satisfy all equations involved. Here, the substitution method is effectively used, which involves substituting one variable into another equation to simplify and find solutions.

Initially, solve one of the equations for a variable. We started with the linear equation:\[ x - y = 1 \]This lets us express \(x\) in terms of \(y\):\[ x = y + 1 \]Then, substitute this expression into the quadratic equation. This substitution transforms the problem into solving a single-variable equation, aiding in isolating solutions efficiently.

Factoring Methods in Solving Equations

The **factoring method** breaks down equations to help solve them, especially handy with quadratic equations. Once substitution made our equation:\[ y^2 + y - 12 = 0 \]We noticed this can be factored into:\[ (y + 4)(y - 3) = 0 \]
This method allows us to set each factor equal to zero, giving potential solutions for \(y\):

• \( y + 4 = 0 \Rightarrow y = -4 \)
• \( y - 3 = 0 \Rightarrow y = 3 \)

Each \(y\)-value is then substituted back into the expression for \(x\) to find the corresponding \(x\)-values. Factoring is effective and simplifies the solution process by breaking equations into bite-sized, solvable components.