To solve the system of equations, we found that y = 28.5 and then substituted this value into the first equation to find x = 8.9375 . The final solution to the system is ( x , y ) = ( 8.9375 , 28.5 ) .
;
Subtract the second equation from the first to eliminate x and solve for y : 4 y = 114 , so y = 28.5 .
Substitute the value of y into the first equation to solve for x : 8 x − 28.5 = 43 , so 8 x = 71.5 .
Solve for x : x = 8 71.5 = 8.9375 .
The solution to the system of equations is x = 8.9375 and y = 28.5 , so the answer is x = 8.9375 , y = 28.5 .
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 equations are:
8 x − y = 43 8 x − 5 y = − 71
We can solve this system using the elimination method.
Eliminate x and solve for y To eliminate x , we can subtract the second equation from the first equation:
( 8 x − y ) − ( 8 x − 5 y ) = 43 − ( − 71 ) 8 x − y − 8 x + 5 y = 43 + 71 4 y = 114
Now, we can solve for y :
y = 4 114 = 2 57 = 28.5
Substitute y and solve for x Now that we have the value of y , we can substitute it back into either of the original equations to solve for x . Let's use the first equation:
8 x − y = 43 8 x − 28.5 = 43 8 x = 43 + 28.5 8 x = 71.5
Now, we can solve for x :
x = 8 71.5 = 16 143 = 8.9375
State the solution Therefore, the solution to the system of equations is x = 8.9375 and y = 28.5 .
Examples
Systems of equations are used in various real-world applications, such as determining the break-even point for a business, calculating the optimal mix of ingredients in a recipe, or modeling supply and demand in economics. For example, a company might use a system of equations to determine the number of units they need to sell to cover their costs and start making a profit. By setting up equations that represent their revenue and expenses, they can solve for the break-even point, which helps them make informed business decisions.