Solve the following linear system of equations:
[tex]
\begin{array}{l}
2 x+y=14 \\
2 x-5 y=2
\end{array}
[/tex]
x =
y =
Answer (2)
Subtract the second equation from the first to eliminate x : ( 2 x + y ) − ( 2 x − 5 y ) = 14 − 2 , which simplifies to 6 y = 12 .
Solve for y : y = 6 12 = 2 .
Substitute y = 2 into the first equation: 2 x + 2 = 14 .
Solve for x : 2 x = 12 , so x = 2 12 = 6 . The solution is x = 6 , y = 2 .
Explanation
Analyzing the problem We are given a system of two linear equations with two variables, x and y :
Equation 1: 2 x + y = 14 Equation 2: 2 x − 5 y = 2
Our goal is to find the values of x and y that satisfy both equations. We can use the method of elimination to solve this system.
Eliminating x To eliminate x , we can subtract Equation 2 from Equation 1:
( 2 x + y ) − ( 2 x − 5 y ) = 14 − 2
Simplifying the left side, we get:
2 x + y − 2 x + 5 y = 6 y
Simplifying the right side, we get:
14 − 2 = 12
So, we have:
6 y = 12
Solving for y Now, we can solve for y by dividing both sides of the equation by 6:
6 6 y = 6 12
y = 2
Substituting y into Equation 1 Now that we have the value of y , we can substitute it back into either Equation 1 or Equation 2 to solve for x . Let's use Equation 1:
2 x + y = 14
Substitute y = 2 :
2 x + 2 = 14
Solving for x Subtract 2 from both sides:
2 x = 14 − 2
2 x = 12
Divide by 2:
x = 2 12
x = 6
Final Answer Therefore, the solution to the system of equations is x = 6 and y = 2 .
Examples
Linear systems of equations are used in various real-world applications, such as determining the optimal mix of products to maximize profit, balancing chemical equations, and analyzing electrical circuits. For example, a company might use a system of equations to determine how many units of two different products they need to produce in order to meet a certain profit target, given the cost of production and the selling price of each product. The solution to the system of equations would provide the number of units of each product that need to be produced.
The solution to the system of equations is x = 6 and y = 2. We found y by eliminating x from the equations and then substituted y back to find x. Both values satisfy the original equations.
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