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Use elimination to solve each system of equations. Check your solution. $$\left\\{\begin{array}{l} 2 x+y=5 \\ 4 x+2 y=3 \end{array}\right.$$

Short Answer

Expert verified
The solution to the system of equations is \( x = -1 \) and \( y = 7 \)

Step by step solution

01

Align Like Terms

We start with the given system of equations: \[ \begin{cases} 2x+y=5 \ 4x+2y=3 \end{cases} \]. It can be noticed that the terms are already aligned.
02

Subtract to eliminate a term

Please notice that by subtracting the second equation from the first one, it would be possible to eliminate the term with \(y\). Do it and we will get \( -2x = 2 \).
03

Find Value of X

Following the above equation \( -2x = 2 \), solving for \(x\) will give \( x = -1 \).
04

Substitute X to find Y

Next, we substitute \(x = -1\) into the first equation of the original system and solve for \(y\). Therefore, \(2*(-1) + y = 5 => y = 5 - (-2) => y = 7\)
05

Check your solution

To assure that the solution is correct, substitute \(x = -1\) and \(y = 7\) into both equations of the original system. It should satisfy both equations. For the first one: \(2*(-1) + 7 = 5\), which holds. For the second one: \(4*(-1) + 2*7 = 3\), which also holds. So, the solution is correct.

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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 strategy to solve a system of linear equations. This method involves adding or subtracting equations to eliminate one of the variables, making it possible to solve for the other variable.

The key to using the elimination method effectively is to manipulate the equations so that adding or subtracting them will result in one variable canceling out. For example, if one equation has the term '2x' and the other has '-2x', adding the two equations will eliminate the 'x' term. Similarly, if the equations have the same coefficient for a variable but with opposite signs, subtraction will eliminate that variable.

To improve the use of the elimination method it's crucial:
  • To align like terms for easy addition or subtraction.
  • To multiply or divide one or both equations to get opposite coefficients for a variable if necessary.
  • To check your solution by plugging the values back into the original equations to ensure they satisfy both equations.
Linear Equations
Linear equations are algebraic expressions where each term is either a constant or the product of a constant and a single variable, and there are no exponents higher than one. These equations can graphically be represented as straight lines, hence the name 'linear'.

A system of linear equations consists of two or more linear equations with the same set of variables. The solution of a linear system is the set of values that satisfies all equations in the system simultaneously. The solution can be a single point where the lines intersect, no point if the lines are parallel and do not meet, or an infinite number of points if the lines coincide.

Understanding the properties of linear equations and their graphs is vital, especially when identifying parallel and coincidental lines, as this affects the number of solutions a system can have.
Substitution Method
The substitution method is another technique used to solve systems of equations. This method entails solving one of the equations for one variable in terms of the others and then substituting this expression into the remaining equations.

This process reduces the number of equations and the number of variables by one. After substituting, the resulting equation is simpler and we can solve for the remaining variable. Once we find the value of one variable, we substitute it back into one of the original equations to find the value of the other variable.

To use the substitution method effectively:
  • Isolate one variable in one equation.
  • Substitute the expression for the isolated variable into the other equation.
  • Solve the resulting single-variable equation.
  • Substitute the value back into the original equation to find the other variable.

It's particularly useful in situations where the coefficients of one of the variables in one of the equations are 1 or -1, making it easier to isolate.
Algebraic Solutions
Algebraic solutions refer to the process of solving equations or systems of equations using algebraic manipulations. This requires a strong understanding of algebraic properties and operations such as addition, subtraction, multiplication, division, and the distributive law.

When solving algebraic equations, the goal is to isolate the variable on one side of the equation to find its value. This involves performing operations that will simplify the equation while keeping it balanced.

In the context of system of equations, algebraic solutions involve finding values for the variables that make all equations in the system true simultaneously. This can be achieved through methods like elimination, substitution, and graphing. It's essential to verify solutions by substituting them back into the original equations to ensure they work for all equations in the system.

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You wish to make a 1 -pound blend of two types of coffee, Kona and Java. The Kona costs \(\$ 8\) per pound and the Java costs \(\$ 5\) per pound. The blend will sell for \(\$ 7\) per pound. (a) Let \(k\) and \(j\) denote the amounts (in pounds) of Kona and Java, respectively, that go into making a 1 -pound blend. One equation that must be satisfied by \(k\) and \(j\) is $$k+j=1$$ Both \(k\) and \(j\) must be between 0 and \(1 .\) Why? (b) Using the variables \(k\) and \(j\), write an equation that expresses the fact that the total cost of 1 pound of the blend will be \(\$ 7\) (c) Solve the system of equations from parts (a) and (b), and interpret your solution. (d) To make a 1 -pound blend of Kona and Java that costs \(\$ 7.50\) per pound, which type of coffee would you use more of? Explain without solving any equations.

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