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Chapter 8: Problem 25

Use elimination to solve each system of equations. Check your solution. $$\left\\{\begin{aligned} -2 x-3 y &=0 \\ 3 x+5 y &=-2 \end{aligned}\right.$$

Short Answer

Expert verified
The solution of the system of equations is x = 6 and y = -4.

Step by step solution

01

Multiply the equations to make the coefficients of y the same

Multiply the first equation by 5 and the second equation by 3 to get: \[\left\{\begin{aligned} -10x-15y & =0 \\ 9x+15y & =-6\end{aligned}\right.\]
02

Employ the elimination method

Add the two new equations together to eliminate y, which results in \(-x=-6\). Solve this equation to find the value of x which is \(x=6\).
03

Substitute the value of x into one of the original equations

Substitute \(x=6\) into the first original equation \(-2x-3y=0\), which results in \(-2(6)-3y=0\). Therefore, we get the equation \(-12-3y=0\) and solving for y gives \(y=-4\).
04

Check the solution

Substitute x=6 and y=-4 into the second original equation and check if the left and right sides are equal. We get \(3(6)+5(-4)\), which simplifies to \(-2\). Thus, the solution checks out since this value equals the right side of the equation.

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Key Concepts

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

Elimination Method
The elimination method is a technique used for solving systems of equations. Its main goal is to simplify the system by combining the equations to cancel out one of the variables. This allows you to solve for the remaining variable more easily. Here's how it works:

First, adjust the coefficients of one variable in both equations so that they are equal but opposite. In our example, we multiplied the first equation by 5 and the second by 3 to align the y terms. This resulted in:
  • Equation 1: \(-10x - 15y = 0\)
  • Equation 2: \(9x + 15y = -6\)
Next, add these equations together. The y variables cancel out, leaving a single equation with just x, making it simple to solve. This method is effective when the equations can be easily manipulated to eliminate a variable.
Solving Equations
Solving equations is about finding the values of the variables that make the equation true. When you have systems of equations, you're looking for values that satisfy all equations at once. In the example, once we eliminated y, we were left with a straightforward equation:
  • \(-x = -6\)
By solving this, we found \(x = 6\). Solving equations often involves basic algebraic steps such as adding, subtracting, multiplying, or dividing both sides by the same number. After finding x, substitute it back into one of the original equations to solve for y. It's crucial to check your solutions by plugging the values back into both original equations to ensure they hold true, confirming the solution is correct.
Substitution Method
The substitution method is another useful technique for solving systems of equations. This approach involves solving one equation for one variable and then substituting that expression into the other equation. Let's see how it works:

After obtaining \(x = 6\) through elimination, substitute x in one of the original equations, such as the first one: \(-2x - 3y = 0\). Plug in \(-2(6)\) to find the equation \(-12 - 3y = 0\).

Solving this gives us \(y = -4\). Substitution is particularly useful when one equation is easy to solve for one of the variables. Even though the textbook solution used elimination primarily, substitution is often a helpful complementary method, especially for checking the consistency of the solution in different scenarios.

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Most popular questions from this chapter

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 \(a=1\)

Consider the following system of equations. $$\left\\{\begin{aligned} x^{2}+y^{2} &=r^{2} \\ (x-h)^{2}+y^{2} &=r^{2} \end{aligned}\right.$$ Let \(r\) be a (fixed) positive number. For what value(s) of \(h\) does this system have (a) exactly one real solution? (b) exactly two real solutions? (c) infinitely many real solutions? (d) no real solution? (Hint: Visualize the graphs of the two equations.)

Sarah can't afford to spend more than \(\$ 90\) per month on transportation to and from work. The bus fare is only \(\$ 1.50\) one way, but it takes Sarah 1 hour and 15 minutes to get to work by bus. If she drives the 15 -mile round trip, her one-way commuting time is reduced to 40 minutes, but it costs her S.40 per mile. If she works at least 20 days a month, how often does she have to drive in order to minimize her commuting time and keep within her monthly budget?

Adult and children's tickets for a certain show sell for \(\$ 8\) each. A total of 1000 tickets are sold, with total sales of \(\$ 8000 .\) Is it possible to figure out exactly how many of each type of ticket were sold? Why or why not?

A family owns and operates three businesses. On their income-tax return, they have to report the depreciation deductions for the three businesses separately. In \(2004,\) their depreciation deductions consisted of use of a car, plus depreciation on 5 -year equipment (on which onc-fifth of the original value is deductible per year) and 10-year equipment (on which one- tenth of the original valuc is deductible per year). The car use (in miles) for cach business in 2004 is given in the following table, along with the original value of the depreciable 5 - and 10 year equipment used in each business that year. $$\begin{array}{|c|c|c|c|}\hline & \begin{array}{c}\text { Car } \\\\\text { Use }\end{array} & \begin{array}{c}\text { Original } \\\\\text { Value, 5-Year }\end{array} & \begin{array}{c}\text { Original } \\\\\text { Value, 10-Year }\end{array} \\\\\text { Business } & \text { (miles) } & \text { Equipment (S) } & \text { Equipment (S) } \\\\\hline 1 & 3200 & 9850 & 435 \\\2 & 8800 & 12,730 & 980 \\\3 & 6880 & 2240 & 615\\\\\hline\end{array}$$ The depreciation deduction for car use in 2004 was 37.5 cents per mile. Use matrix multiplication to determine the total depreciation deduction for each business in 2004.
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