Solve For X And Y 0.2 X + 0.3 Y Is Equal To 1 And 4x + 0.5 Y Is Equal To 2.3 Solve By Substitution Method
Introduction
In algebra, a system of linear equations is a set of two or more equations that contain two or more variables. Solving a system of linear equations involves finding the values of the variables that satisfy all the equations in the system. In this article, we will solve a system of two linear equations using the substitution method. The two equations are:
0.2x + 0.3y = 1 4x + 0.5y = 2.3
Understanding the Substitution Method
The substitution method is a technique used to solve a system of linear equations. It involves solving one equation for one variable and then substituting that expression into the other equation. This process is repeated until the values of the variables are found.
Step 1: Solve the First Equation for x
Let's solve the first equation for x:
0.2x + 0.3y = 1
Subtract 0.3y from both sides:
0.2x = 1 - 0.3y
Divide both sides by 0.2:
x = (1 - 0.3y) / 0.2
Step 2: Substitute the Expression for x into the Second Equation
Now, substitute the expression for x into the second equation:
4x + 0.5y = 2.3
Substitute x = (1 - 0.3y) / 0.2:
4((1 - 0.3y) / 0.2) + 0.5y = 2.3
Simplifying the Equation
Multiply both sides by 0.2 to eliminate the fraction:
4(1 - 0.3y) + 0.1y = 0.46
Expand the equation:
4 - 1.2y + 0.1y = 0.46
Combine like terms:
3.8y = 0.46 - 4
Simplify the equation:
3.8y = -3.54
Step 3: Solve for y
Divide both sides by 3.8:
y = -3.54 / 3.8
y = -0.93
Step 4: Substitute the Value of y into One of the Original Equations
Now, substitute the value of y into one of the original equations to find the value of x. Let's use the first equation:
0.2x + 0.3y = 1
Substitute y = -0.93:
0.2x + 0.3(-0.93) = 1
Simplify the equation:
0.2x - 0.279 = 1
Add 0.279 to both sides:
0.2x = 1.279
Divide both sides by 0.2:
x = 1.279 / 0.2
x = 6.395
Conclusion
In this article, we solved a system of two linear equations using the substitution method. We first solved one equation for x and then substituted that expression into the other equation. We simplified the resulting equation and solved for y. Finally, we substituted the value of y into one of the original equations to find the value of x. The values of x and y are x = 6.395 and y = -0.93.
Example Use Cases
The substitution method can be used to solve a wide range of systems of linear equations. Here are a few example use cases:
- Finance: A company has two investment options, A and B. Option A has a return of 0.2x + 0.3y, where x is the amount invested in stocks and y is the amount invested in bonds. Option B has a return of 4x + 0.5y. The company wants to find the optimal investment strategy that maximizes returns. Using the substitution method, we can solve for x and y to find the optimal investment strategy.
- Science: A scientist is studying the relationship between two variables, x and y. The scientist has two equations that describe the relationship between x and y: 0.2x + 0.3y = 1 and 4x + 0.5y = 2.3. Using the substitution method, we can solve for x and y to find the values of the variables that satisfy both equations.
- Engineering: An engineer is designing a system that involves two variables, x and y. The engineer has two equations that describe the relationship between x and y: 0.2x + 0.3y = 1 and 4x + 0.5y = 2.3. Using the substitution method, we can solve for x and y to find the values of the variables that satisfy both equations.
Conclusion
In conclusion, the substitution method is a powerful technique for solving systems of linear equations. It involves solving one equation for one variable and then substituting that expression into the other equation. This process is repeated until the values of the variables are found. The substitution method can be used to solve a wide range of systems of linear equations, including finance, science, and engineering applications.