With a one unit change in 'b' what is the change in 'y' in the solution of the system of equations x + y = 2; 1.01x + 0.99y = b?

System of Equations Analysis

We are provided with a system of two linear equations:

1. $x+y=2$
2. $1.01x+0.99y=b$

The goal is to determine the change in $y$ that occurs when there is a one-unit change in the variable $b$. To achieve this, we need to express $y$ as a function of $b$ and then find its rate of change with respect to $b$, which is represented by the derivative $\frac{dy}{db}$.

Solving for Y in Terms of B

To establish the relationship between $y$ and $b$, we will solve this system of equations for $y$. The substitution method is an effective way to do this.

From the first equation, $x+y=2$, we can easily express $x$ in terms of $y$:

- $x=2-y$

Now, we substitute this expression for $x$ into the second equation of the system:

- $1.01(2-y)+0.99y=b$

Next, we distribute the $1.01$ and simplify the equation by combining like terms:

- $2.02-1.01y+0.99y=b$

Combine the terms that contain $y$:

- $2.02-(1.01-0.99)y=b$
- $2.02-0.02y=b$

Now, we isolate the term with $y$ and then solve for $y$ to express it in terms of $b$:

- $-0.02y=b-2.02$
- $y=\frac{b-2.02}{-0.02}$
- $y=\frac{b}{-0.02}-\frac{2.02}{-0.02}$
- $y=-50b-(-101)$
- $y=-50b+101$

So, we have successfully expressed $y$ as a linear function of $b$.

Change in Y with Respect to B

To find the change in $y$ for a one-unit change in $b$, we need to determine the derivative of $y$ with respect to $b$, which is $\frac{dy}{db}$. Since the relationship $y=-50b+101$ is a linear equation, the change in $y$ for every one-unit change in $b$ is simply the coefficient of $b$.

- $\frac{dy}{db}=\frac{d}{db}(-50b+101)$
- $\frac{dy}{db}=-50$

This result indicates that for every one-unit increase in $b$, the value of $y$ will decrease by $50$ units. Therefore, the change in $y$ for a one unit change in $b$ is $-50$.

Summary of Calculation StepsThus, the change in $y$ for a one-unit change in $b$ is $-50$.