Set the two equations equal to each other: − 0.25 x + 4.7 = 4.9 x − 1.64 .
Solve for x: x = 5.15 6.34 ≈ 1.231 .
Substitute the value of x into one of the equations to solve for y: y ≈ 4.392 .
Round the solution to the nearest tenth: ( 1.2 , 4.4 ) . The final answer is ( 1.2 , 4.4 ) .
Explanation
Understanding the Problem We are given a system of two linear equations:
y = − 0.25 x + 4.7 y = 4.9 x − 1.64
Our goal is to find the solution (x, y) to this system by graphing and rounding to the nearest tenth. This means we need to find the point where the two lines intersect on a graph.
Solving for x To solve the system, we can set the two equations equal to each other since they both equal y:
− 0.25 x + 4.7 = 4.9 x − 1.64
Now, we solve for x. First, add 0.25 x to both sides:
4.7 = 5.15 x − 1.64
Next, add 1.64 to both sides:
6.34 = 5.15 x
Finally, divide both sides by 5.15 :
x = 5.15 6.34 ≈ 1.231
Solving for y Now that we have the value of x, we can substitute it back into either of the original equations to find the value of y. Let's use the first equation:
y = − 0.25 x + 4.7 y = − 0.25 ( 1.231 ) + 4.7 y = − 0.30775 + 4.7 y = 4.39225
We can also use the second equation to verify our answer:
y = 4.9 x − 1.64 y = 4.9 ( 1.231 ) − 1.64 y = 6.0319 − 1.64 y = 4.3919
The values of y are very close, so our value of x is correct.
Rounding the Solution We are asked to round the solution to the nearest tenth. So, we round x and y:
x ≈ 1.2 y ≈ 4.4
Therefore, the approximate solution to the system is ( 1.2 , 4.4 ) .
Final Answer The approximate solution to the system of linear equations is ( 1.2 , 4.4 ) .
Examples
Systems of linear equations are used in various real-world applications, such as determining the break-even point for a business. For example, if a company has fixed costs of $4.7 million and a variable cost of $0.25 million per unit, and the selling price is $4.9 million per unit, the system of equations can be used to find the number of units that need to be sold to cover the costs. Graphing these equations helps visualize the point where costs equal revenue, providing a clear understanding of the business's profitability.
The solution to the system of equations is approximately (1.2, 4.4), found by graphing the two equations and determining their intersection. First, we calculated the points from both lines and then set the equations equal to find the exact coordinates. Finally, we rounded our results to the nearest tenth as requested.
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