Problem 17 - Precalculus with Limits

Question

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

Step-by-Step Solution

Answer

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

Step 1: 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\(4x + 10y = 6\).

Step 2: Eliminating a Variable

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

Step 3: Solving for the Remaining Variable

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

Step 4: 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:\(3x + 2y = 10\)\(3*4 + 2y = 10\)\(12 + 2y = 10\)\(2y = 10 -12 = -2\)\(y = \frac{-2}{2} = -1\).

Step 5: 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.

Key Concepts

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.