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Chapter 12: Problem 36

Solve each system. Use any method you wish. $$ \left\\{\begin{array}{r} x^{2}-3 y^{2}+1=0 \\ 2 x^{2}-7 y^{2}+5=0 \end{array}\right. $$

Short Answer

Expert verified
(2鈭�2, 鈭�3), (-2鈭�2, 鈭�3), (2鈭�2, -鈭�3), (-2鈭�2, -鈭�3)

Step by step solution

01

Analyze the System of Equations

We have two equations in the system: 1) \(x^2 - 3y^2 + 1 = 0\) 2) \(2x^2 - 7y^2 + 5 = 0\)
02

Isolate One Equation

Isolate \(x^2\) in the first equation: \(x^2 = 3y^2 - 1\)
03

Substitute into Second Equation

Substitute \(x^2 = 3y^2 - 1\) into the second equation: \(2(3y^2 - 1) - 7y^2 + 5 = 0\)
04

Simplify the Substituted Equation

Simplify: \(6y^2 - 2 - 7y^2 + 5 = 0\)\(-y^2 + 3 = 0\)\(y^2 = 3\)
05

Solve for y

Take the square root on both sides: \(y = \pm\sqrt{3}\)
06

Solve for x

Substitute \(y = \pm\sqrt{3}\) back into \(x^2 = 3y^2 - 1\): For \(y = \sqrt{3}\): \(x^2 = 3(\sqrt{3})^2 - 1 = 3(3) - 1 = 9 - 1 = 8\)\(x = \pm\sqrt{8} = \pm2\sqrt{2}\)For \(y = -\sqrt{3}\): The result will be the same since \(y^2 = 3\), so \(x = \pm2\sqrt{2}\)
07

Write the Final Solutions

The solutions to the system are: \((2\sqrt{2}, \sqrt{3})\), \((-2\sqrt{2}, \sqrt{3})\), \((2\sqrt{2}, -\sqrt{3})\), \((-2\sqrt{2}, -\sqrt{3})\)

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Key Concepts

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

Understanding Algebraic Elimination
Algebraic elimination is a method used for solving systems of equations. It focuses on eliminating one variable by combining equations. This method helps simplify complex problems. Let's break down the steps:
  • **Identify the system**: Start with a set of equations.
  • **Isolate a variable**: Look for a way to make one equation simpler by isolating one variable.
  • **Substitute and combine**: Substitute the isolated variable into the other equation, which now has only one variable.
  • **Solve for the remaining variable**: With one variable eliminated, solve the remaining equation.
  • **Back-substitute to find the eliminated variable**: Use the value obtained to find the other variable.
In our solution, we isolated \(x^2\) in the equation \(x^2 - 3y^2 + 1 = 0\) and substituted it into the other equation. By combining and simplifying, we eliminated \(x\) first to solve for \(y\). Then, we substituted back to find \(x\). This approach can be powerful, especially in solving linear systems and can also be adapted for nonlinear systems like quadratics.
Exploring the Substitution Method
The substitution method is another great technique to solve systems of equations. It is particularly useful when one of the equations is easy to manipulate. Here are the steps:
  • **Solve one equation for a variable**: Pick the easier looking equation, and isolate one variable.
  • **Substitute**: Replace the isolated variable in the other equation to express it in terms of a single variable.
  • **Solve the new equation**: Now that you have an equation with one variable, solve it.
  • **Substitute back**: Finally, substitute the found value back into the isolated variable equation.
In our exercise, we solved for \(x^2\) from the first equation: \(x^2 = 3y^2 - 1\). Then, we substituted this into the second equation. This converted the system into a single equation \(6y^2 - 2 - 7y^2 + 5 = 0\) in terms of \(y\). Solving for \(y\) became straightforward. Finally, we substituted back to find \(x\).
Solving Quadratic Equations
Solving quadratic equations is a crucial algebraic skill. Quadratic equations take the form \(ax^2 + bx + c = 0\). Here's how to solve them:
  • **Factoring**: Express the quadratic as a product of two binomials; solve for the roots.
  • **Using the Quadratic Formula**: If factoring is tricky, use the formula \(x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\).
  • **Completing the Square**: Transform the quadratic into a perfect square trinomial. Then solve for the roots.
In our problem, we simplified the equation to \(y^2 = 3\). The solutions were \(y = \pm\sqrt{3}\). For substitution, we solved for \(x\) by plugging \(y\) back, resulting in \(x = \pm2\sqrt{2}\). The solutions were obtained by basic knowledge of square roots.

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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. $$ \left\\{\begin{array}{rr} x-2 y+3 z= & 7 \\ 2 x+y+z= & 4 \\ -3 x+2 y-2 z= & -10 \end{array}\right. $$

A young couple has \(\$ 25,000\) to invest. As their financial consultant, you recommend that they invest some money in Treasury bills that yield \(7 \%,\) some money in corporate bonds that yield \(9 \%,\) and some money in junk bonds that yield \(11 \% .\) Prepare a table showing the various ways that this couple can achieve the following goals: (a) \(\$ 1500\) per year in income (b) \(\$ 2000\) per year in income (c) \(\$ 2500\) per year in income (d) What advice would you give this couple regarding the income that they require and the choices available?

The length of fence required to enclose a rectangular field is 3000 meters. What are the dimensions of the field if it is known that the difference between its length and width is 50 meters?

A doctor's prescription calls for the creation of pills that contain 12 units of vitamin \(\mathrm{B}_{12}\) and 12 units of vitamin E. Your pharmacy stocks three powders that can be used to make these pills: one contains \(20 \%\) vitamin \(\mathrm{B}_{12}\) and \(30 \%\) vitamin \(\mathrm{E} ;\) a second, \(40 \%\) vitamin \(\mathrm{B}_{12}\) and \(20 \%\) vitamin \(\mathrm{E}\) and a third, \(30 \%\) vitamin \(\mathrm{B}_{12}\) and \(40 \%\) vitamin \(\mathrm{E}\). Create \(\mathrm{a}\) table showing the possible combinations of these powders that could be mixed in each pill. Hint: 10 units of the first powder contains \(10 \cdot 0.2=2\) units of vitamin \(\mathrm{B}_{12}\).

Verify that the values of the variables listed are solutions of the system of equations. $$ \begin{array}{l} \left\\{\begin{array}{l} 4 x-z=7 \\ 8 x+5 y-z=0 \\ -x-y+5 z=6 \\ \end{array}\right.\\\ x=2, y=-3, z=1 \\ (2,-3,1) \end{array} $$
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