Rewrite the second equation to isolate y : y = 5 x − 7.4 .
Substitute the expression for y into the first equation and simplify: 2 x + 3 ( 5 x − 7.4 ) = 16.9 ⇒ 17 x = 39.1 .
Solve for x : x = 17 39.1 = 2.3 .
Substitute the value of x back into the equation y = 5 x − 7.4 to find y : y = 5 ( 2.3 ) − 7.4 = 4.1 . The solution is ( 2.3 , 4.1 ) .
Explanation
Analyze the problem We are given the following system of linear equations:
2 x + 3 y = 16.9 5 x = y + 7.4
Our goal is to find the values of x and y that satisfy both equations. We will solve this system by algebraic manipulation and substitution.
Isolate y in the second equation First, let's rewrite the second equation to isolate y :
5 x = y + 7.4 y = 5 x − 7.4
Substitute into the first equation Now, substitute the expression for y from the second equation into the first equation:
2 x + 3 ( 5 x − 7.4 ) = 16.9
Simplify the equation Expand and simplify the equation:
2 x + 15 x − 22.2 = 16.9 17 x = 16.9 + 22.2 17 x = 39.1
Solve for x Solve for x :
x = 17 39.1 x = 2.3
Solve for y Now that we have the value of x , substitute it back into the equation y = 5 x − 7.4 to find the value of y :
y = 5 ( 2.3 ) − 7.4 y = 11.5 − 7.4 y = 4.1
State the solution Therefore, the solution to the system of equations is x = 2.3 and y = 4.1 .
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 linear equations to find the number of units they need to sell to cover their costs and start making a profit. The solution to this system would give the break-even point, which is a crucial piece of information for business planning and decision-making. In this case, the equations might represent the cost and revenue functions, and solving the system would tell us the quantity at which cost equals revenue.