Value: 3
Solve the system:
[tex]\begin{array}{l}
y=-3 x \
x+y=4\end{array}[/tex]
a. (-2,6)
b. (-4,12)
c. (-1,3)
d. (1,-3)
Answer (2)
Substitute y = − 3 x into x + y = 4 .
Simplify to get − 2 x = 4 , so x = − 2 .
Substitute x = − 2 into y = − 3 x to find y = 6 .
The solution is ( − 2 , 6 ) .
Explanation
Understanding the Problem We are given a system of two equations with two variables, x and y . The equations are:
y = − 3 x
x + y = 4
We need to find the values of x and y that satisfy both equations. The possible solutions are given as options a, b, c, and d.
Substitution We can use the substitution method to solve this system. Substitute the first equation, y = − 3 x , into the second equation:
x + ( − 3 x ) = 4
Simplifying the Equation Simplify the equation:
− 2 x = 4
Solving for x Solve for x by dividing both sides by -2:
x = − 2 4 = − 2
Solving for y Substitute the value of x = − 2 back into the first equation to find y :
y = − 3 ( − 2 ) = 6
Checking the Solution The solution is x = − 2 and y = 6 . This corresponds to the point ( − 2 , 6 ) . Now, we check if this solution matches any of the given options.
The solution ( − 2 , 6 ) matches option a.
Examples
Systems of equations are used in various real-life scenarios, such as determining the break-even point for a business. For example, if a company's cost function is y = 5 x + 1000 (where y is the total cost and x is the number of units produced) and the revenue function is y = 15 x , solving this system of equations will give the number of units the company needs to sell to break even. This concept is also used in physics to solve problems involving multiple forces or velocities.
The solution to the system of equations is ( − 2 , 6 ) , which corresponds to option a. We found this by substituting y = − 3 x into the other equation and solving for both variables. The final result was confirmed by checking both original equations.
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