\left\{\begin{array}{l}36 x-11 y=-14 \\ 24 x-17 y=10\end{array}\right.
Answer (2)
Multiply the first equation by 24 and the second equation by 36 to prepare for eliminating x .
Subtract the modified equations to eliminate x and solve for y , finding y = − 2 .
Substitute y = − 2 into one of the original equations and solve for x , resulting in x = − 1 .
The solution to the system of equations is { x = − 1 y = − 2 .
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:
{ 36 x − 11 y = − 14 24 x − 17 y = 10
Eliminate x To solve this system of equations, we can use the method of elimination. We aim to eliminate one of the variables by multiplying the equations by suitable constants so that the coefficients of either x or y become equal or additive inverses. Let's eliminate x . To do this, we can multiply the first equation by 24 and the second equation by 36.
Multiplying the first equation by 24, we get:
24 ( 36 x − 11 y ) = 24 ( − 14 ) 864 x − 264 y = − 336
Multiplying the second equation by 36, we get:
36 ( 24 x − 17 y ) = 36 ( 10 ) 864 x − 612 y = 360
Solve for y Now we have the following system of equations:
{ 864 x − 264 y = − 336 864 x − 612 y = 360
Subtract the first equation from the second equation to eliminate x :
( 864 x − 612 y ) − ( 864 x − 264 y ) = 360 − ( − 336 ) 864 x − 612 y − 864 x + 264 y = 360 + 336 − 348 y = 696
Calculate y Now, solve for y :
y = − 348 696 y = − 2
Substitute y into the first equation Substitute the value of y back into one of the original equations to solve for x . Let's use the first equation:
36 x − 11 y = − 14 36 x − 11 ( − 2 ) = − 14 36 x + 22 = − 14 36 x = − 14 − 22 36 x = − 36
Calculate x Now, solve for x :
x = 36 − 36 x = − 1
State the solution Therefore, the solution to the system of equations is x = − 1 and y = − 2 .
{ x = − 1 y = − 2
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 a product. The cost to produce the product includes a fixed cost (e.g., rent) and a variable cost (e.g., materials). The revenue from selling the product depends on the price and the number of units sold. By setting up a system of equations, the company can determine the number of units they need to sell to cover their costs and start making a profit. This is a crucial aspect of business planning and financial analysis.
Let's say the fixed costs are $10,000, and the variable cost per unit is $5. The selling price per unit is $10. We can set up the following equations:
Total Cost: C = 10000 + 5 x Total Revenue: R = 10 x
To find the break-even point, we set C = R :
10000 + 5 x = 10 x 10000 = 5 x x = 2000
So, the company needs to sell 2000 units to break even. This example demonstrates how systems of equations help in making informed business decisions.
To solve the system of equations, we used elimination to find y = − 2 and then substituted back to find x = − 1 . The final solution is x = − 1 and y = − 2 .
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