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Chapter 12: Problem 32

Solve each system. Use any method you wish. $$ \left\\{\begin{array}{r} 2 y^{2}-3 x y+6 y+2 x+4=0 \\ 2 x-3 y+4=0 \end{array}\right. $$

Short Answer

Expert verified
Solutions are \( \left( \frac{7+3\sqrt{41}}{10}, \frac{9 + \sqrt{41}}{5} \right) \) and \( \left( \frac{7-3\sqrt{41}}{10}, \frac{9 - \sqrt{41}}{5} \right) \)

Step by step solution

01

Solve the Linear Equation for x

Start by isolating one variable in the linear equation. The second equation is linear: \[ 2x - 3y + 4 = 0 \] Solving for \(x\), we get: \[ x = \frac{3y - 4}{2} \]
02

Substitute x into the Quadratic Equation

Substitute the expression for \(x\) from Step 1 into the quadratic equation: \[ 2y^2 - 3xy + 6y + 2x + 4 = 0 \] Replace \(x\) with \( \frac{3y-4}{2} \): \[ 2y^2 - 3y \left( \frac{3y-4}{2} \right) + 6y + 2 \left( \frac{3y-4}{2} \right) + 4 = 0 \]
03

Simplify the Equation

Simplify the equation by combining like terms and eliminating fractions. Start by distributing and simplifying: \[ 2y^2 - \frac{9y^2 - 12y}{2} + 6y + 3y - 4 + 4 = 0 \] Combine like terms: \[ 2y^2 - \frac{9y^2}{2} + 9y - 4 = 0 \] Multiply through by 2 to clear the fraction: \[ 4y^2 - 9y^2 + 18y - 8 = 0 \] Combining like terms: \[ -5y^2 + 18y - 8 = 0 \]
04

Solve the Quadratic Equation

Solve the quadratic equation using the quadratic formula \(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = -5\), \(b = 18\), and \(c = -8\): \[ y = \frac{-18 \pm \sqrt{18^2 - 4(-5)(-8)}}{2(-5)} \] Calculate the discriminant: \[ y = \frac{-18 \pm \sqrt{324 - 160}}{-10} \] Simplify further: \[ y = \frac{-18 \pm \sqrt{164}}{-10} \] Finally: \[ y = \frac{-18 \pm 2\sqrt{41}}{-10} \] Simplifying yields two possible values: \[ y = \frac{9}{5} \pm \frac{\sqrt{41}}{5} \]
05

Find Corresponding x Values

Use the values of \(y\) found in Step 4 to solve for \(x\) using \( x = \frac{3y - 4}{2} \). For each value of \(y\): For \( y_1 = \frac{9 + \sqrt{41}}{5} \): \[ x = \frac{3 \left( \frac{9 + \sqrt{41}}{5} \right) - 4}{2} = \frac{3(9 + \sqrt{41}) - 20}{10} = \frac{27 + 3\sqrt{41} - 20}{10} = \frac{7 + 3\sqrt{41}}{10} \] For \( y_2 = \frac{9 - \sqrt{41}}{5} \): \[ x = \frac{3 \left( \frac{9 - \sqrt{41}}{5} \right) - 4}{2} = \frac{3(9 - \sqrt{41}) - 20}{10} = \frac{27 - 3\sqrt{41} - 20}{10} = \frac{7 - 3\sqrt{41}}{10} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

linear equations
Linear equations are algebraic expressions that represent straight lines when graphed on a coordinate plane. They have the general form: \[ ax + by + c = 0 \ \] where \(a\), \(b\), and \(c\) are constants. In our exercise, the linear equation is \[ 2x - 3y + 4 = 0 \]. To solve a system involving a linear equation, one usually aims to isolate one of the variables, as done in Step 1:
quadratic equations
Quadratic equations are polynomial equations of degree 2, generally written 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. In our exercise, the quadratic equation is: \[ 2y^2 - 3xy + 6y + 2x + 4 = 0 \ \] These equations can often have two solutions, as they represent positions on the parabola. Understanding how to manipulate and solve these equations is crucial for tackling the given problem:
substitution method
The substitution method is a technique for solving systems of equations, where one solves for one variable in terms of another using one equation and substitutes this expression into the other equation. In our exercise:
  • Step 1: Solve the linear equation for one variable.
  • Step 2: Substitute this expression into the quadratic equation.
This method simplifies a system by reducing it to a single equation in one variable.
quadratic formula
The quadratic formula is used to find the roots of a quadratic equation and is given by:\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \ \]. The discriminant (\ \sqrt{b^2 - 4ac}) determines the nature of the roots:
  • If positive, there are two real and distinct solutions.
  • If zero, there is one real solution.
  • If negative, the solutions are complex numbers.
In our problem, we used the quadratic formula to solve the reduced quadratic equation:
\[-5y^2 + 18y - 8 = 0 \ \].

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Most popular questions from this chapter

Solve each system of equations. If the system has no solution, state that it is inconsistent. $$ \left\\{\begin{array}{rr} x+4 y-3 z= & -8 \\ 3 x-y+3 z= & 12 \\ x+y+6 z= & 1 \end{array}\right. $$

Solve each system of equations. If the system has no solution, state that it is inconsistent. $$ \left\\{\begin{array}{r} x+2 y-z=-3 \\ 2 x-4 y+z=-7 \\ -2 x+2 y-3 z=4 \end{array}\right. $$

Curve Fitting Find real numbers \(a, b,\) and \(c\) so that the graph of the function \(y=a x^{2}+b x+c\) contains the points \((-1,-2),(1,-4),\) and (2,4)

Solve each system of equations. If the system has no solution, state that it is inconsistent. $$ \left\\{\begin{aligned} \frac{1}{2} x+y &=-2 \\ x-2 y &=8 \end{aligned}\right. $$

Solve each system of equations. If the system has no solution, state that it is inconsistent. $$ \left\\{\begin{array}{rr} x-y-z= & 1 \\ -x+2 y-3 z= & -4 \\ 3 x-2 y-7 z= & 0 \end{array}\right. $$
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