Solve this system of equations using the elimination method.
[tex]\begin{array}{l}
3 x-2 y=0 \\
2 x+y=7
\end{array}[/tex]
The solution to the system is ( [ ] , [ ] ).
Answer (2)
By applying the elimination method to the system of equations, we find that x = 2 and y = 3 . The complete solution is therefore ( 2 , 3 ) , which indicates the values that satisfy both equations simultaneously.
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Multiply the second equation by 2: 4 x + 2 y = 14 .
Add the modified second equation to the first equation: 7 x = 14 .
Solve for x : x = 2 .
Substitute x = 2 into 2 x + y = 7 and solve for y : y = 3 . The solution is ( 2 , 3 ) .
Explanation
Analyze the problem We are given the following system of equations:
{ 3 x − 2 y = 0 2 x + y = 7
Our goal is to solve for x and y using the elimination method.
Multiply the second equation by 2 To eliminate y , we can multiply the second equation by 2:
2 ( 2 x + y ) = 2 ( 7 )
This simplifies to:
4 x + 2 y = 14
Add the equations Now, we add the modified second equation to the first equation:
( 3 x − 2 y ) + ( 4 x + 2 y ) = 0 + 14
This simplifies to:
7 x = 14
Solve for x Now, we solve for x :
x = 7 14 = 2
Solve for y Substitute the value of x back into the second original equation to solve for y :
2 ( 2 ) + y = 7
4 + y = 7
y = 7 − 4 = 3
State the solution Therefore, the solution to the system of equations is ( 2 , 3 ) .
Examples
Systems of equations are used in various real-world 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. In this case, imagine you're trying to balance two different nutritional components in a meal to meet specific dietary requirements. Each equation represents a constraint on the amounts of these components, and solving the system helps you find the exact quantities needed to satisfy both requirements simultaneously.
