Solve the following system of equations algebraically: [tex]$30 x+44 y=15$[/tex] [tex]$40 x+55 y=13$[/tex]
Answer (1)
Multiply the first equation by 4 and the second equation by 3 to make the coefficients of x equal: 120 x + 176 y = 60 and 120 x + 165 y = 39 .
Subtract the second equation from the first to eliminate x: 11 y = 21 .
Solve for y: y = 11 21 .
Substitute y back into the first original equation and solve for x: x = 10 − 23 .
The solution is x = 10 − 23 , y = 11 21 .
Explanation
Analyze the problem We are given a system of two linear equations in two variables x and y:
Equation 1: 30 x + 44 y = 15 Equation 2: 40 x + 55 y = 13
Our objective is to solve this system of equations algebraically to find the values of x and y that satisfy both equations.
Prepare for elimination To solve the system, we can use the method of elimination. We'll multiply each equation by a constant so that the coefficients of either x or y are the same (or negatives of each other). Let's eliminate x.
Multiply Equation 1 by 4: 4 ( 30 x + 44 y ) = 4 ( 15 ) which simplifies to 120 x + 176 y = 60 .
Multiply Equation 2 by 3: 3 ( 40 x + 55 y ) = 3 ( 13 ) which simplifies to 120 x + 165 y = 39 .
Eliminate x Now we have two new equations:
Equation 3: 120 x + 176 y = 60 Equation 4: 120 x + 165 y = 39
Subtract Equation 4 from Equation 3 to eliminate x: ( 120 x + 176 y ) − ( 120 x + 165 y ) = 60 − 39 11 y = 21
Solve for y Solve for y: y = 11 21
Solve for x Substitute the value of y back into one of the original equations to solve for x. Let's use Equation 1: 30 x + 44 ( 11 21 ) = 15 30 x + 4 ( 21 ) = 15 30 x + 84 = 15 30 x = 15 − 84 30 x = − 69 x = 30 − 69 = 10 − 23
State the solution So the solution to the system of equations is x = 10 − 23 and y = 11 21 .
Verify the solution To verify the solution, substitute the values of x and y into both original equations:
Equation 1: 30 ( 10 − 23 ) + 44 ( 11 21 ) = 15 − 69 + 84 = 15 15 = 15 (Correct)
Equation 2: 40 ( 10 − 23 ) + 55 ( 11 21 ) = 13 − 92 + 105 = 13 13 = 13 (Correct)
The solution is verified.
Final Answer The solution to the system of equations is:
x = 10 − 23 y = 11 21
Examples
Systems of equations are used in many real-world applications, such as determining the break-even point for a business. For example, a company might use a system of equations to model its costs and revenues, and then solve the system to find the level of production at which its costs equal its revenues. This is a crucial calculation for making informed business decisions. Another example is in electrical engineering, where systems of equations are used to analyze circuits and determine the currents and voltages at different points in the circuit. These applications highlight the importance of understanding how to solve systems of equations.
