Solve the system of linear equations by graphing. Round the solution to the nearest tenth as needed.
[tex]\begin{array}{l}
y+2.3=0.45 x \\
-2 y=4.2 x-7.8
\end{array}[/tex]
Answer (2)
Rewrite the given equations in slope-intercept form: y = 0.45 x − 2.3 and y = − 2.1 x + 3.9 .
Set the two equations equal to each other: 0.45 x − 2.3 = − 2.1 x + 3.9 .
Solve for x : x ≈ − 1.2 .
Substitute the value of x back into one of the equations to solve for y : y ≈ 2.4 . The solution is ( − 1.2 , 2.4 ) .
Explanation
Analyze the problem and convert to slope-intercept form We are given a system of two linear equations:
y + 2.3 = 0.45 x
− 2 y = 4.2 x − 7.8
Our goal is to solve this system by graphing, which means finding the point where the two lines intersect. We need to express each equation in slope-intercept form ( y = m x + b ) to make them easier to graph.
Rewrite the first equation Let's rewrite the first equation in slope-intercept form:
y + 2.3 = 0.45 x
Subtract 2.3 from both sides:
y = 0.45 x − 2.3
Rewrite the second equation Now, let's rewrite the second equation in slope-intercept form:
− 2 y = 4.2 x − 7.8
Divide both sides by -2:
y = − 2.1 x + 3.9
Set the equations equal Now we have the two equations in slope-intercept form:
y = 0.45 x − 2.3
y = − 2.1 x + 3.9
To find the solution, we need to find the point of intersection. We can set the two equations equal to each other:
0.45 x − 2.3 = − 2.1 x + 3.9
Solve for x Solve for x :
0.45 x + 2.1 x = 3.9 + 2.3
2.55 x = 6.2
x = 2.55 6.2 ≈ − 1.2058823529411755
Solve for y Now, substitute the value of x back into either equation to find y . Let's use the first equation:
y = 0.45 ( − 1.2058823529411755 ) − 2.3
y = − 0.542647058823529 − 2.3
y = 2.431372549019607
Find the solution So the solution is approximately x = − 1.2058823529411755 and y = 2.431372549019607 . Rounding to the nearest tenth, we get x ≈ − 1.2 and y ≈ 2.4 .
Final Answer Therefore, the solution to the system of equations, rounded to the nearest tenth, is ( − 1.2 , 2.4 ) .
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 and variable costs per unit, and they sell each unit at a certain price, we can set up a system of equations to find the number of units they need to sell to cover their costs and start making a profit. Another example is in mixture problems, where we need to find the amounts of different solutions to mix to obtain a desired concentration.
The solution to the system of equations is approximated to be (-1.2, 2.4). This point represents where the two lines intersect on a graph, showing the values of x and y that satisfy both equations. To find this solution, both equations were rewritten in slope-intercept form and then solved for their intersection point.
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