Solve the following system of equations algebraically: [tex]$30 x+44 y=1$[/tex] [tex]$40 x+55 y=13$[/tex].
Answer (1)
Multiply the first equation by 4 and the second equation by 3 to prepare for eliminating x .
Subtract the modified equations to eliminate x and solve for y : y = − 11 35 .
Substitute the value of y back into one of the original equations and solve for x : x = 10 47 .
The solution to the system of equations is x = 10 47 and y = − 11 35 , which we write as x = 10 47 , y = − 11 35 .
Explanation
Analyze the problem We are given a system of two linear equations with two variables, x and y . Our goal is to find the values of x and y that satisfy both equations simultaneously. The given equations are:
30 x + 44 y = 1 40 x + 55 y = 13
Eliminate x To solve this system of equations, we can use the method of elimination. We'll multiply each equation by a suitable constant so that the coefficients of either x or y in both equations become equal or additive inverses. This will allow us to eliminate one variable when we add or subtract the equations.
Multiply the first equation by 4 and the second equation by 3:
4 ( 30 x + 44 y ) = 4 ( 1 ) ⟹ 120 x + 176 y = 4 3 ( 40 x + 55 y ) = 3 ( 13 ) ⟹ 120 x + 165 y = 39
Solve for y Now, subtract the second equation from the first to eliminate x :
( 120 x + 176 y ) − ( 120 x + 165 y ) = 4 − 39 11 y = − 35
Calculate y Solve for y :
y = 11 − 35
Substitute y into the first equation Substitute the value of y back into the first original equation to solve for x :
30 x + 44 ( − 11 35 ) = 1 30 x − 4 ( 35 ) = 1 30 x − 140 = 1
Calculate x Solve for x :
30 x = 141 x = 30 141 = 10 47
State the solution Therefore, the solution to the system of equations is:
x = 10 47 , y = − 11 35
Examples
Systems of equations are used in various real-world applications, such as determining the break-even point for a business. For example, suppose a company produces and sells widgets. The cost to produce x widgets is given by C = 5 x + 1000 , where 5 x represents the variable cost and 1000 represents the fixed costs. The revenue from selling x widgets is given by R = 15 x . The break-even point is where the cost equals the revenue, so we have the system of equations:
C = 5 x + 1000 R = 15 x
Setting C = R , we get 5 x + 1000 = 15 x , which simplifies to 10 x = 1000 , and thus x = 100 . So, the company needs to produce and sell 100 widgets to break even. This example demonstrates how systems of equations can be used to model and solve real-world problems in business and economics.
