Solve the second equation for a : a = 11 − 2 c .
Substitute this expression into the first equation: 2 ( 11 − 2 c ) − 3 c = − 6 .
Solve for c : c = 4 .
Substitute the value of c back into the equation for a : a = 3 . The solution is ( 3 , 4 ) .
Explanation
Analyze the problem We are given a system of two linear equations with two variables, a and c . Our goal is to find the values of a and c that satisfy both equations simultaneously. The given equations are:
2 a − 3 c = − 6 a + 2 c = 11
We can solve this system using either substitution or elimination. Here, we will use the substitution method.
Solve for a First, we solve the second equation for a in terms of c :
a = 11 − 2 c
Substitute into first equation Next, we substitute this expression for a into the first equation:
2 ( 11 − 2 c ) − 3 c = − 6
Solve for c Now, we simplify and solve for c :
22 − 4 c − 3 c = − 6 22 − 7 c = − 6 − 7 c = − 6 − 22 − 7 c = − 28 c = − 7 − 28 c = 4
Solve for a Now that we have the value of c , we substitute it back into the equation a = 11 − 2 c to find the value of a :
a = 11 − 2 ( 4 ) a = 11 − 8 a = 3
Final Answer Therefore, the solution to the system of equations is a = 3 and c = 4 . We can write this as an ordered pair ( a , c ) = ( 3 , 4 ) .
Examples
Systems of linear equations are used in various real-world applications, such as determining the optimal mix of products in manufacturing, balancing chemical equations, and modeling supply and demand in economics. For instance, a company might use a system of equations to determine how many units of two different products they need to sell to reach a specific revenue target, given the prices of the products and their production costs. By solving the system, they can find the exact quantities of each product needed to meet their goal, ensuring efficient resource allocation and maximizing profits.