$\begin{aligned} 3 y+11 & =4 x \\ 10 x+2 y+1 & =0 \end{aligned}$
Answer (1)
Rearrange the equations to the standard form: − 4 x + 3 y = − 11 and 10 x + 2 y = − 1 .
Eliminate x by multiplying the first equation by 2 5 and adding it to the second equation, resulting in 2 19 y = − 2 57 .
Solve for y : y = − 3 .
Substitute y = − 3 into one of the equations to solve for x : x = 2 1 .
The solution is x = 2 1 , y = − 3 .
Explanation
Understanding 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.
Equation 1: 3 y + 11 = 4 x Equation 2: 10 x + 2 y + 1 = 0
Rearranging the Equations First, let's rearrange the equations into the standard form A x + B y = C .
Equation 1: 3 y + 11 = 4 x can be rewritten as − 4 x + 3 y = − 11 .
Equation 2: 10 x + 2 y + 1 = 0 can be rewritten as 10 x + 2 y = − 1 .
Setting up for Elimination Now we have the following system of equations:
− 4 x + 3 y = − 11 10 x + 2 y = − 1
We can solve this system using either the substitution or elimination method. Let's use the elimination method.
Multiplying Equation 1 To eliminate x , we can multiply the first equation by 2 5 to make the coefficients of x opposites:
2 5 ∗ ( − 4 x + 3 y ) = 2 5 ∗ ( − 11 ) − 10 x + 2 15 y = − 2 55
Adding the Equations Now we add the modified Equation 1 to Equation 2 to eliminate x :
( − 10 x + 2 15 y ) + ( 10 x + 2 y ) = − 2 55 + ( − 1 ) 2 15 y + 2 y = − 2 55 − 2 2 2 15 y + 2 4 y = − 2 57 2 19 y = − 2 57
Solving for y Now, solve for y :
y = 2 19 − 2 57 y = − 2 57 ∗ 19 2 y = − 19 57 y = − 3
Solving for x Substitute the value of y back into either Equation 1 or Equation 2 to solve for x . Let's use Equation 1:
− 4 x + 3 y = − 11 − 4 x + 3 ( − 3 ) = − 11 − 4 x − 9 = − 11 − 4 x = − 11 + 9 − 4 x = − 2 x = − 4 − 2 x = 2 1
Verifying the Solution Therefore, the solution to the system of equations is x = 2 1 and y = − 3 .
Let's verify the solution by substituting the values of x and y into both original equations.
Equation 1: 3 y + 11 = 4 x 3 ( − 3 ) + 11 = 4 ( 2 1 ) − 9 + 11 = 2 2 = 2
Equation 2: 10 x + 2 y + 1 = 0 10 ( 2 1 ) + 2 ( − 3 ) + 1 = 0 5 − 6 + 1 = 0 0 = 0
The solution satisfies both equations.
Final Answer The solution to the system of equations is:
x = 2 1 y = − 3
Examples
Simultaneous equations are used in various real-life scenarios, 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, simultaneous equations can be used to find the number of units they need to sell to cover their costs and start making a profit. They are also used in physics to solve problems involving multiple forces or constraints, and in computer graphics to perform transformations and projections.
