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Chapter 7: Problem 17

Solving a System by Elimination In Exercises \(13-30,\) solve the system by the method of elimination and check any solutions algebraically. $$ \left\\{\begin{array}{l}{3 x+2 y=10} \\ {2 x+5 y=3}\end{array}\right. $$

Short Answer

Expert verified
The solution to the system of equations is \(x = 4, y = -1\).

Step by step solution

01

Rearrangement

Start by multiplying the first equation by 5 and the second equation by 2 in order to allow for the y variables to cancel out when subtracted. This gives:\(15x + 10y = 50\) and\(4x + 10y = 6\).
02

Eliminating a Variable

Subtract the second equation from the first to eliminate the y variable:\(15x - 4x = 50 - 6\)This simplifies down to:\(11x = 44\).
03

Solving for the Remaining Variable

Now, solving for x is straightforward:\(x = \frac{44}{11} = 4\).
04

Solving for the Eliminated Variable

Now, substitute \(x = 4\) back into either of the original equations to solve for y. Here, we use the first original equation:\(3x + 2y = 10\)\(3*4 + 2y = 10\)\(12 + 2y = 10\)\(2y = 10 -12 = -2\)\(y = \frac{-2}{2} = -1\).
05

Checking the Solutions

Finally, substitute \(x = 4\) and \(y = -1\) into both original equations to check. This gives:Into the first equation:\(3*4 + 2*(-1) = 10\), which simplifies to 12 - 2 = 10, thus holds true.And the second equation:\(2*4 + 5*(-1) = 3\), which simplifies to 8 - 5 = 3, thus also holds true.

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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 powerful technique for solving systems of linear equations. It involves eliminating one variable so that you can solve for the other. This method is particularly useful when the equations are lined up in standard form. Here is how it typically works:
  • First, you may need to manipulate the equations by multiplying them so that the coefficients of one variable are the same.
  • Once the coefficients match, you add or subtract the equations to eliminate that variable.
  • This leaves you with a single equation in one variable.
In our problem, both equations were manipulated such that the y-coefficients become equal. Subtracting them from each other allowed us to eliminate the y variable, simplifying the process of finding x.
Solving Systems Algebraically
Solving systems algebraically means finding the values of the variables using algebraic methods. Unlike graphing, algebraic methods provide precise results and are efficient for more complex systems. The elimination method, as used in our example, is a form of algebraic solving:
  • It involved restructuring the equations to make the coefficients of one variable equal.
  • By eliminating y, we directly found the value of x.
  • Once one variable is known, substitution can be used in any of the original equations to find the other variable.
This process not only reveals the solution but also enhances understanding of the relationship between the two equations.
Linear Equations
Linear equations are equations of the first order. They are graphs of straight lines when plotted and can often be written in the form of: ax + by = c, where a, b, and c are constants. In the system provided:
  • The first equation is 3x + 2y = 10, representing one line.
  • The second equation is 2x + 5y = 3, representing another line.
Our goal here is to find the point at which these two lines intersect, as that point (x, y) is the solution to the system. These equations have only one solution as they are not parallel or identical.
Checking Solutions
Once the potential solutions are found, it is essential to verify their accuracy. This involves plugging the values back into the original equations to ensure they satisfy both equations. In our example:
  • The solution found was x = 4 and y = -1.
  • Substituting into the first equation 3(4) + 2(-1) = 10 checks out, confirming this equation holds true at this solution.
  • Similarly, substituting into the second equation 2(4) + 5(-1) = 3 also confirms its validity.
This step is critical for ensuring that the solution is correct and consistent for both equations. It reinforces confidence in the solution's accuracy.

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Most popular questions from this chapter

Think About It Are the following two systems of equations equivalent? Give reasons for your answer. $$ \left\\{ \begin{aligned} x + 3 y - z & = 6 \\ 2 x - y + 2 z & = 1 \\ 3 x + 2 y - z & = 2 \end{aligned} \right. $$ $$ \left\\{ \begin{aligned} x + 3 y - z = & 6 \\ - 7 y + 4 z = & 1 \\ - 7 y - 4 z = & \- 16 \end{aligned} \right. $$

Solving a Linear Programming Problem, sketch the region determined by the constraints. Then find the minimum and maximum values of the objective function (if possible) and where they occur, subject to the indicated constraints. $$ \begin{array}{c}{\text { Objective function: }} \\ {z=5 x+\frac{1}{2} y} \\\ {\text { Constraints: }} \\ {x \geq 0} \\ {y \geq 0} \\ {\frac{1}{2} x+y \leq 8} \\ {x+\frac{1}{2} y \geq 4}\end{array} $$

Solving a Linear Programming Problem, find the minimum and maximum values of the objective function and where they occur, subject to the indicated constraints. (For each exercise, the graph of the region determined by the constraints is provided.) $$ \begin{array}{c}{\text { Objective function: }} \\ {z=2 x+5 y} \\ {\text { Constraints: }} \\ {x \geq 0} \\ {y \geq 0} \\ {x+3 y \leq 15} \\ {4 x+y \leq 16}\end{array} $$

Prescriptions The numbers of prescriptions \(P\) (in thousands) filled at two pharmacies from 2009 through 2013 are shown in the table. $$ \begin{array}{|c|c|c|}\hline \text { Year } & {\text { Pharmacy A }} & {\text { Pharmacy } \mathrm{B}} \\ \hline 2009 & {19.2} & {20.4} \\ \hline 2010 & {19.6} & {20.8} \\ \hline 2011 & {20.0} & {21.1} \\ \hline 2012 & {20.6} & {21.5} \\ \hline 2013 & {21.3} & {22.0} \\ \hline\end{array} $$ (a) Use a graphing utility to create a scatter plot of the data for pharmacy A and find a linear model. Let \(t\) represent the year, with \(t=9\) corresponding to \(2009 .\) Repeat the procedure for pharmacy B. (b) Assuming that the numbers for the given five years are representative of future years, will the number of prescriptions filled at pharmacy A ever exceed the number of prescriptions filled at pharmacy B? If so, then when?

Advanced Applications In Exercises 73 and \(74 ,\) find values of \(x , y ,\) and \(\lambda\) that satisfy the system. These systems arise in certain optimization problems in calculus, and \(\lambda\) is called a Lagrange multiplier. $$ \left\\{ \begin{array} { r } { 2 x - 2 x \lambda = 0 } \\ { - 2 y + \lambda = 0 } \\ { y - x ^ { 2 } = 0 } \end{array} \right. $$
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