/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 8 Solving a System by Elimination ... [FREE SOLUTION] | 91Ó°ÊÓ

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Solving a System by Elimination In Exercises \(5-12,\) solve the system by the method of elimination. Label each line with its equation. To print an enlarged copy of the graph, go to MathGraphs.com. $$ \left\\{\begin{array}{l}{2 x-y=-3} \\ {4 x+3 y=-21}\end{array}\right. $$

Short Answer

Expert verified
The solution to this system of equations is \(x = -3, y = -3\)

Step by step solution

01

Rewrite the system of equations

The given system of equations is \[\begin{align*}2x - y &= -3 \hspace{3cm} (1) \4x + 3y &= -21 \hspace{3cm} (2)\end{align*}\]
02

Multiply Equation (1) by 3

To get rid of \(y\), we should multiply equation (1) by 3 which is the coefficient of \(y\) in equation (2):\[\begin{align*}6x - 3y &= -9 \hspace{3cm} (3) \4x + 3y &= -21 \hspace{3cm} (4)\end{align*}\]
03

Adding Equation (3) and Equation (4)

This will eliminate \(y\) since \(-3y\) and \(+3y\) cancels out each other and we get a new equation with one variable:\[\begin{align*}(6x - 3y) + (4x + 3y) &= (-9) + (-21) \10x &= -30\end{align*}\]
04

Solving for x

We divide both sides by 10 to isolate \(x\):\[\begin{align*}\frac{10x}{10} &= \frac{-30}{10} \x &= -3\end{align*}\]
05

Substituting x in Equation (1)

Now that we know the value of \(x\), we can substitute it back into Equation (1) to find the value of \(y\):\[\begin{align*}2(-3) - y &= -3 \-6 - y &= -3 \-y &= -3 + 6 \-y &= 3 \y &= -3\end{align*}\]

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Key Concepts

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

The Elimination Method
Understanding the elimination method is crucial for solving systems of linear equations. It's essentially a technique to reduce a system of equations to one equation with one variable. With this method, you typically alter the equations in the system to achieve oppositely signed terms that cancel each other out when added together.

To use the elimination method effectively, follow these steps: identify a term in both equations with the same or opposite coefficients, multiply one or both equations, if necessary, to obtain such terms, and then add or subtract the equations to eliminate one of the variables.

In the provided exercise, we multiplied Equation (1) by 3 to match the coefficient of 'y' in Equation (2). This strategic move set the stage for us to then add the two equations together, eliminating the variable 'y' and paving the way to solve for 'x'. Once 'x' is found, it can be substituted back into one of the original equations to find 'y', completing the solution to the system.

Improvement Advice: When employing this method, always double-check your calculations, ensure your coefficients are lined up correctly before adding or subtracting, and consider the simplicity of numbers when choosing which variable to eliminate. This can save time and reduce the risk of error.
Understanding Systems of Linear Equations
A system of linear equations consists of two or more linear equations with the same set of variables. The goal when solving a system is to find the values of the variables that satisfy all equations simultaneously.

These equations can graphically represent lines on a coordinate plane, and the solution to the system corresponds to the point or points where the lines intersect. Depending on the slopes and y-intercepts, systems may have one solution (intersect at one point), no solution (lines are parallel and never intersect), or infinitely many solutions (lines coincide).

Systems like the one in our exercise, with one solution, are called independent and consistent. It's crucial to write the equations clearly, label them for easy reference, and proceed systematically through the solution process. Remember that finding the solution to a system is about exploring the relationship between the variables and distilling this relationship down to specific numeric values.
Algebraic Substitution
Algebraic substitution is a foundational technique that's especially handy when you've simplified a system of equations to a single equation with one variable. You solve for this 'easy' variable and then 'substitute' the obtained value back into one of the original equations to find the other variable.

Substitution ensures that the solution obtained is indeed valid for both original equations. It's a methodical approach: solve one equation, use its result in the other, and solve the resulting simpler equation. Substitution is like the puzzle piece that links together the steps in solving a system of equations.

We applied algebraic substitution in Step 5 of the solution by taking the value of 'x' and inserting it into Equation (1) to solve for 'y'. This final step is a crucial verification to ensure the solution is correct and fits both equations.

Improvement Advice: Always simplify the equation as much as possible before substituting, and follow through with the arithmetic carefully to avoid sign errors, which are common pitfalls in this stage.

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Most popular questions from this chapter

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