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In Exercises \(1-34\), find all real solutions of the system of equations. If no real solution exists, so state. $$\left\\{\begin{array}{r} 3 x^{2}-10 y=5 \\ x-y=-2 \end{array}\right.$$

Short Answer

Expert verified
The real solutions to the system of equations are \( (x_{1}, y_{1}) = (\frac{11 + \sqrt{37}}{3} - 2, \frac{11 + \sqrt{37}}{3}) \) and \( (x_{2}, y_{2}) = (\frac{11 - \sqrt{37}}{3} - 2, \frac{11 - \sqrt{37}}{3} ) \)

Step by step solution

01

Express one variable in terms of the other in the second equation

In the equation \(x - y = -2\), you can express \(x\) in terms of \(y\): \(x = y - 2\)
02

Substitute into the first equation

Substitute \(x = y - 2\) into the first equation: \(3(y - 2)^2 - 10y = 5\)
03

Simplify the equation

Now we have a quadratic equation in terms of \(y\). Simplify to get: \(3(y^2 - 4y + 4) - 10y = 5\), and further simplify this to \(3y^2 - 12y + 12 - 10y = 5\), which simplifies to \(3y^2 - 22y + 12 = 5\), and further simplification gives \(3y^2 - 22y + 7 = 0\)
04

Solve for variable

Now solve for the variable \(y\) by solving quadratic equation \(3y^2 - 22y + 7 = 0\). This gives \(y = \frac{22 \pm \sqrt{(-22)^2 - 4*3*7}}{2*3}\), which simplifies to \(y1 = \frac{11 + \sqrt{37}}{3}\) and \(y2 = \frac{11 - \sqrt{37}}{3}\)
05

Back-substitute to find x

Substitute these found values of y back into the one of the original equations to find the corresponding values of x. We have, using the equation \(x = y - 2\), \( x_{1} = y_{1} - 2 = \frac{11 + \sqrt{37}}{3} - 2 \) and \( x_{2} = y_{2} - 2 = \frac{11 - \sqrt{37}}{3} - 2\)
06

Verify solutions

You could check at this point if these solutions make both equations true. If they do, these are the correct solutions.

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Key Concepts

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

Quadratic Equations
Quadratic equations are a type of polynomial equation where the highest exponent of the variable is 2. They typically take the form \(ax^2 + bx + c = 0\). These equations are significant in mathematics because they can model a wide range of phenomena, from projectile motion to optimization problems.
In the context of the exercise, we come across a quadratic equation when we substitute \(x = y - 2\) into the equation \(3x^2 - 10y = 5\). After completing the substitution and simplifying, the equation is transformed into a quadratic in terms of \(y\):
\[3y^2 - 22y + 7 = 0\]
An essential feature of quadratic equations is that they can have either two, one, or no real solutions, depending on the discriminant \(\Delta = b^2 - 4ac\):
  • If \(\Delta > 0\), there are two distinct real solutions.
  • If \(\Delta = 0\), there is exactly one real solution.
  • If \(\Delta < 0\), there are no real solutions.
For our equation, the discriminant helps us determine how many real solutions \(y\) has, guiding us to find the corresponding \(x\) values.
Substitution Method
The substitution method is a straightforward approach for solving systems of equations, especially when one of the equations can be easily solved for one of the variables. It involves replacing one variable with an equivalent expression from another equation.
To tackle the problem in the exercise, we start with the system:
  • \(3x^2 - 10y = 5\)
  • \(x - y = -2\)
By expressing \(x\) in terms of \(y\) from the second equation, \(x = y - 2\), we find a direct relation between the two variables. We then substitute this expression for \(x\) into the first equation:
\[3(y-2)^2 - 10y = 5\]
This technique simplifies the original system to a single equation with one variable, \(y\), which can be solved using methods specific to quadratic equations. Once values for \(y\) are found, they can be substituted back into the expression for \(x\) to find the corresponding \(x\) values.
Real Solutions
In mathematical terms, a real solution is a solution to an equation that is a real number, as opposed to a complex or imaginary number. Determining whether a system of equations has real solutions often involves evaluating the discriminant of a quadratic equation.
For a quadratic equation \(ax^2 + bx + c = 0\), its discriminant \(\Delta = b^2 - 4ac\) reveals the nature of the roots:
  • \(\Delta > 0\): Two distinct real solutions.
  • \(\Delta = 0\): One real solution (also called a repeated or double root).
  • \(\Delta < 0\): No real solutions (the solutions are complex or imaginary).
In this exercise, we calculated the discriminant of the quadratic \(3y^2 - 22y + 7 = 0\) and found it to be positive. Thus, the equation has two real solutions for \(y\). These solutions are then used to find corresponding \(x\) values, giving us two solution pairs \((x_1, y_1)\) and \((x_2, y_2)\) for the system of equations. It is crucial to verify that these solutions satisfy both original equations, confirming their validity.

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Most popular questions from this chapter

An electronics firm makes a clock radio in two different models: one (model 380 ) with a battery backup feature and the other (model 360 ) without. It takes 1 hour and 15 minutes to manufacture each unit of the model 380 radio, and only 1 hour to manufacture each unit of the model \(360 .\) At least 500 units of the model 360 radio are to be produced. The manufacturer realizes a profit per radio of \(\$ 15\) for the model 380 and only \(\$ 10\) for the model \(360 .\) If at most 2000 hours are to be allocated to the manufacture of the two models combined, how many of each model should be made to maximize the total profit?

For what value(s) of \(b\) does the following system of equations have two distinct, real solutions? $$\left\\{\begin{array}{l} y=-x^{2}+2 \\ y=x+b \end{array}\right.$$

This set of exercises will draw on the ideas presented in this section and your general math background. Find the inverse of $$\left[\begin{array}{lll}a & a & a \\\0 & 1 & 0 \\\0 & 0 & 1\end{array}\right]$$ where \(a\) is nonzero. Evaluate this inverse for the case in which \(a=1\)

Apply elementary row operations to a matrix to solve the system of equations. If there is no solution, state that the system is inconsistent. $$\left\\{\begin{array}{c}x+2 y+z=-3 \\ 3 x+y-2 z=2 \\ 4 x+3 y-z=0\end{array}\right.$$

Involve the use of matrix multiplication to transform one or more points. This technique, which can be applied to any set of points, is used extensively in computer graphics. Let \(A=\left[\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right]\) and \(B=\left[\begin{array}{r}2 \\ -1\end{array}\right]\) (a) Calculate the product matrix \(A B\) (b) On a single coordinate system, plot the point (2,-1) and the point whose coordinates \((x, y)\) are the entries of the product matrix found in part (a). Explain geometrically what the matrix multiplication did to the point (2,-1) (c) How would you undo the multiplication in part (a)?

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