$\begin{array}{l}3 x+4 y=-3 \\ f y=\frac{-x-1}{2}\end{array}$

Question

03-11-2026
Mathematics

Answered

$\begin{array}{l}3 x+4 y=-3 \\ f y=\frac{-x-1}{2}\end{array}$

Verified Answer :

Related Questions

Homophones a a female sheep an evergreen tree b sandy shore kind of tree c guided a metal d opens lock harbour e flat land joiners tool f suffering piece of gla

A compact disc player cost 50 plus 17 12 VAT how much does the compact disc player cost

Gravitational elastic and chemical are three forms of

what is national library week

What is the product 6437 2368 2368 1184 2048

describe the process by which DNA is uses to make proteins

3x2y7 2x5y10 how to solve by elimination

Which is true of the Mexican Revolution a It was won by Native Americans and poor farmers b It was won by foreigners c It was won by conservatives d It was won

King Henry VIII is most closely associated with A the invention of the printing press B the expansion of Englands territory C the translation of the Bible into

the organization of congress in to committees is important because of

GinnyAnswer GinnyAnswer · Answer 1 · 2026-11-03T21:26:26-05:00

We rewrite the second equation to solve for $x$ in terms of $y$ and $f$: $x = -2fy - 1$.
Substituting this into the first equation leads to the simplified equation $y(4 - 6f) = 0$.
If $f eq \frac{2}{3}$, the only solution is $y = 0$, which gives $x = -1$.
If $f = \frac{2}{3}$, the system is dependent and has infinitely many solutions of the form $\left(-\frac{4}{3}y - 1, y\right)$.
Final Answer: $\boxed{f eq \frac{2}{3}: (-1, 0); f = \frac{2}{3}: \left(-\frac{4}{3}y - 1, y\right)}$

Explanation

Analyzing the System of Equations

We are given a system of two linear equations in variables $x$ and $y$, with $f$ acting as a parameter:

$3x + 4y = -3$
$fy = \frac{-x - 1}{2}$

Our goal is to solve for $x$ and $y$ in terms of the parameter $f$. We will use the method of substitution to eliminate one of the variables.

Substitution and Simplification

We begin by isolating one variable in the second equation. Multiply both sides of the second equation by 2 to clear the fraction:
\[2fy = -x - 1\]
Next, solve for $x$:
\[x = -2fy - 1\]
Now, substitute this expression for $x$ into the first equation:
\[3(-2fy - 1) + 4y = -3\]

Solving for y

Expand the substituted equation:
\[-6fy - 3 + 4y = -3\]
Group the terms involving $y$:
\[-6fy + 4y - 3 = -3\]
Add 3 to both sides to simplify:
\[-6fy + 4y = 0\]
Factor out $y$:
\[y(4 - 6f) = 0\]
This equation shows that the product of $y$ and $(4 - 6f)$ must be zero. This gives us two distinct scenarios based on the value of the parameter $f$.

Analyzing Cases for Parameter f

Case 1: If $4 - 6f eq 0$, which is equivalent to $f eq \frac{2}{3}$, then the only solution for $y$ is $y = 0$.
Substituting $y = 0$ back into our expression for $x$:
\[x = -2f(0) - 1 = -1\]
Thus, for $f eq \frac{2}{3}$, the unique solution is $(x, y) = (-1, 0)$.

Case 2: If $f = \frac{2}{3}$, then the equation becomes $y(0) = 0$, which is true for any real number $y$.
Substituting $f = \frac{2}{3}$ back into the expression $x = -2fy - 1$:
\[x = -2\left(\frac{2}{3}\right)y - 1 = -\frac{4}{3}y - 1\]
In this case, there are infinitely many solutions of the form $\left(-\frac{4}{3}y - 1, y\right)$ for any real number $y$.

Conclusion

In summary, we have solved the system of equations. For $f eq \frac{2}{3}$, the solution is $\boxed{(-1, 0)}$. For $f = \frac{2}{3}$, the solutions are $\boxed{\left(-\frac{4}{3}y - 1, y\right)}$ for any $y \in \mathbb{R}$.

Examples

This type of system of equations appears frequently in engineering when analyzing control systems. The parameter $f$ might represent a gain or feedback value. When $f$ is not at a critical value (in this case $\frac{2}{3}$), the system reaches a single steady state. When $f$ hits that critical point, the system might become unstable or have an infinite set of equilibrium points, which is vital for engineers to understand to design safe and robust systems.