\begin{array}{l}3 w-4 v=62 \\ 2 w-v=8\end{array}
Answer (2)
Solve the second equation for v : v = 2 w − 8 .
Substitute this expression for v into the first equation and solve for w : w = − 6 .
Substitute the value of w back into the expression for v and solve for v : v = − 20 .
The solution to the system of equations is w = − 6 , v = − 20 .
Explanation
Problem Analysis We are given a system of two linear equations with two variables, w and v . Our goal is to find the values of w and v that satisfy both equations simultaneously.
Given Equations We have the following system of equations:
Equation 1: 3 w − 4 v = 62
Equation 2: 2 w − v = 8
Solving for v We can solve this system using the substitution or elimination method. Let's use the substitution method. From Equation 2, we can express v in terms of w :
v = 2 w − 8
Substitution Now, substitute this expression for v into Equation 1:
3 w − 4 ( 2 w − 8 ) = 62
Solving for w Simplify and solve for w :
3 w − 8 w + 32 = 62
− 5 w = 30
w = − 6
Solving for v Now that we have the value of w , we can substitute it back into the expression for v :
v = 2 ( − 6 ) − 8
v = − 12 − 8
v = − 20
Solution So, the solution to the system of equations is w = − 6 and v = − 20 . Let's verify this solution by substituting these values back into the original equations.
Verification Equation 1: 3 w − 4 v = 62
3 ( − 6 ) − 4 ( − 20 ) = − 18 + 80 = 62 (Correct)
Equation 2: 2 w − v = 8
2 ( − 6 ) − ( − 20 ) = − 12 + 20 = 8 (Correct)
Both equations are satisfied, so our solution is correct.
Final Answer Therefore, the solution to the system of equations is:
w = − 6
v = − 20
Examples
Systems of equations are used in various real-life applications, such as determining the break-even point for a business, calculating the optimal mix of ingredients in a recipe, or modeling supply and demand in economics. For instance, if a company sells two products, we can set up a system of equations to determine the number of units of each product that need to be sold to reach a certain profit target. By solving the system, we can find the values of the variables (number of units) that satisfy the profit equation and other constraints, helping the company make informed decisions about production and sales.
The solution to the system of equations is w = − 6 and v = − 20 , which can be verified by substituting these values back into the original equations. The substitution method was used to isolate one variable and solve for the other. Each equation was verified to confirm the accuracy of the solution.
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