Problem 17 - Beginning and Intermediate Algebra

Question

Solve each system. $$\begin{array}{c} 3 x^{2}+4 y=-1 \\ x^{2}+3 y=-12 \end{array}$$

Step-by-Step Solution

Answer

To solve the given system of equations, first isolate \(x^2\) from the second equation: \(x^2 = -3y - 12\). Substitute this into the first equation: \(3(-3y - 12) + 4y = -1\). Simplify and solve for \(y\): \(y=-7\). Substitute \(y\) back into the second equation to solve for \(x\): \(x^2 - 21 = -12\). Solve for \(x\): \(x = ±3\). Therefore, there are two possible solutions: \((x, y) = (3, -7)\) and \((x, y) = (-3, -7)\).

Step 1: Isolate x² from the second equation

From the second equation (x² + 3y = -12), we can isolate x² by subtracting 3y from both sides of the equation: x² = -3y - 12

Step 2: Substitute x² into the first equation

Now, we can substitute the expression for x² from Step 1 into the first equation (3x² + 4y = -1): 3(-3y - 12) + 4y = -1

Step 3: Solve for y

Now, let's simplify and solve for y by first distributing the 3: -9y - 36 + 4y = -1 Combine like terms: -5y - 36 = -1 Now, add 36 to both sides of the equation: -5y = 35 Finally, divide both sides by -5 to find the value of y: y = -7

Step 4: Substitute y back into the second equation

With the value of y found in Step 3, we can substitute it back into the second equation (x² + 3y = -12) to find the value of x: x² + 3(-7) = -12

Step 5: Solve for x

Now, let's solve for x: x² - 21 = -12 Add 21 to both sides of the equation: x² = 9 Now, we find the square root of both sides to obtain the values of x: x = ±3

Step 6: Present the solutions

Finally, we have two possible solutions, as there are two different values of x: Solution 1: (x, y) = (3, -7) Solution 2: (x, y) = (-3, -7)

Key Concepts

Quadratic Equations

Quadratic equations are a central part of algebra and are commonly represented in the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants. These equations form a parabola when graphed on a coordinate plane. Quadratic equations can have either two distinct solutions, one solution (where both solutions are equal), or no real solutions, depending on the discriminant (\(b^2 - 4ac\)). In the given problem,

• we have two equations, one of which is quadratic in nature: \(3x^2 + 4y = -1\) and \(x^2 + 3y = -12\).
• When solving these systems, we're interested in finding the values of \(x\) and \(y\) that satisfy both equations simultaneously.

Understanding the characteristics of quadratic equations helps in realizing why the solution might involve two or more values for a variable, as seen in the original exercise.

Substitution Method

The substitution method is a systematic approach for solving systems of equations. It involves solving one of the equations for one variable and then substituting that expression into the other equations. To illustrate:

• The first step in the solution was to express \(x^2\) from the second equation \(x^2 + 3y = -12\) as \(x^2 = -3y - 12\).
• Following this, we substitute \(x^2\) in the first equation, \(3x^2 + 4y = -1\).

By substituting \(x^2\) with \(-3y - 12\), we reduce the first equation to a simpler form involving only \(y\). This reduction simplifies the problem to a single variable equation, which can be easier to solve.

Solving for Variables

Once substitution has been carried out, we need to solve for the variables. This process requires isolating the variable of interest, often by using basic algebraic operations. In the given problem, after substitution, the equation became :

• \(-9y - 36 + 4y = -1\)
• Simplifying this, it becomes \(-5y - 36 = -1\).

Next, adding 36 to both sides reduces the equation further, \(-5y = 35\). Finally, dividing both sides by \(-5\), we find \(y = -7\).With \(y\) determined, the step back involves substituting this value back into one of the original equations to find \(x\). This back substitution is vital for finding any remaining unknowns, ensuring that solutions satisfy both parts of the system.

Algebraic Solutions

Algebraic solutions to equations involve calculating exact solutions through manipulation and simplification of the expressions. Unlike numerical methods, which approximate, algebraic solutions provide the precise values of the variables. For the given quadratic system:

• We determined \(y = -7\) through a series of algebraic manipulations, which involved combining like terms and performing operations symmetrically on both sides of the equation.
• The solution for \(x\) required solving \(x^2 - 21 = -12\), giving \(x^2 = 9\).
• Taking the square root of both sides results in \(x = ±3\).

Thus, the pair of solutions \((x, y) = (3, -7)\) and \((-3, -7)\) were found, demonstrating the completeness of the algebraic solution method in capturing all potential solutions for \(x\).