\begin{aligned} 4 x+3 y &=-14 \\ 8 x+10 y &=-36\end{aligned}
Answer (2)
Multiply the first equation by -2: − 8 x − 6 y = 28 .
Add the modified equation to the second equation: 4 y = − 8 .
Solve for y : y = − 2 .
Substitute y = − 2 into the first equation and solve for x : x = − 2 .
The solution to the system of equations is x = − 2 , y = − 2 .
Explanation
Understanding the Problem We are given a system of two linear equations in two variables, x and y . Our goal is to find the values of x and y that satisfy both equations simultaneously.
Eliminating x To solve this system, we can use the method of elimination. This involves manipulating the equations so that when we add or subtract them, one of the variables is eliminated. Let's multiply the first equation by -2: − 2 ( 4 x + 3 y ) = − 2 ( − 14 ) − 8 x − 6 y = 28
Solving for y Now, we add the modified first equation to the second equation: ( − 8 x − 6 y ) + ( 8 x + 10 y ) = 28 + ( − 36 ) − 8 x + 8 x − 6 y + 10 y = − 8 4 y = − 8
Finding the value of y Divide both sides by 4 to solve for y : 4 4 y = 4 − 8 y = − 2
Substituting y into the first equation 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: 4 x + 3 y = − 14 4 x + 3 ( − 2 ) = − 14 4 x − 6 = − 14
Isolating x Add 6 to both sides: 4 x − 6 + 6 = − 14 + 6 4 x = − 8
Finding the value of x Divide both sides by 4 to solve for x : 4 4 x = 4 − 8 x = − 2
Final Answer Therefore, the solution to the system of equations is x = − 2 and y = − 2 .
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 traffic flow in a city. For example, a company might use a system of equations to determine how many units of a product they need to sell to cover their costs and start making a profit. By setting up equations that represent the company's revenue and expenses, they can solve for the number of units that will result in a break-even point.
To solve the given system of equations, we used the elimination method. After manipulating the equations, we found that y = − 2 and substituted this back into the first equation to discover x = − 2 . Therefore, the solution is x = − 2 and y = − 2 .
;
