Consider the equation and the following ordered pairs: $(2, y)$ and $(x,-3)$.
[tex]4 x-2 y=2[/tex]
Step 1 of 2: Compute the missing [tex]x[/tex] and [tex]y[/tex] values so that each ordered pair will satisfy the given equation.
Answer (1)
Substitute x = 2 into the equation 4 x − 2 y = 2 and solve for y , which gives y = 3 .
Substitute y = − 3 into the equation 4 x − 2 y = 2 and solve for x , which gives x = − 1 .
The missing values are y = 3 and x = − 1 .
The ordered pairs that satisfy the equation are ( 2 , 3 ) and ( − 1 , − 3 ) .
x = − 1 , y = 3
Explanation
Understanding the Problem We are given the equation 4 x − 2 y = 2 and two ordered pairs ( 2 , y ) and ( x , − 3 ) . We need to find the missing x and y values so that each ordered pair satisfies the given equation.
Finding the Value of y First, let's find the value of y when x = 2 . Substitute x = 2 into the equation 4 x − 2 y = 2 :
4 ( 2 ) − 2 y = 2 8 − 2 y = 2 Subtract 8 from both sides: − 2 y = 2 − 8 − 2 y = − 6 Divide both sides by -2: y = − 2 − 6 y = 3 So, the first ordered pair is ( 2 , 3 ) .
Finding the Value of x Next, let's find the value of x when y = − 3 . Substitute y = − 3 into the equation 4 x − 2 y = 2 :
4 x − 2 ( − 3 ) = 2 4 x + 6 = 2 Subtract 6 from both sides: 4 x = 2 − 6 4 x = − 4 Divide both sides by 4: x = 4 − 4 x = − 1 So, the second ordered pair is ( − 1 , − 3 ) .
Final Answer Therefore, the missing values are y = 3 and x = − 1 .
Examples
Understanding how to solve for variables in linear equations is a fundamental concept in algebra and has numerous real-world applications. For instance, consider a scenario where a company's profit is linearly related to the number of products sold. If the relationship is expressed as P = a x + b , where P is the profit, x is the number of products sold, and a and b are constants, you can use ordered pairs of (products sold, profit) to determine the constants a and b . This allows the company to predict future profits based on sales or determine the number of products needed to reach a specific profit target. Linear equations are also used in physics to describe motion with constant velocity, in economics to model supply and demand, and in everyday situations like calculating the cost of a taxi ride based on distance traveled.
