Linear Equations with Two Variables: Using fractions

To solve this problem, we'll follow these steps:

• Step 1: Solve the first equation for $x$.
• Step 2: Substitute this expression for $x$ into the second equation.
• Step 3: Analyze the resulting equation for $y$.
• Step 4: Determine the relationship between the two equations.

Now, let's work through each step:
Step 1: Starting with the first equation $\frac{x}{2} + y = 3$, we solve for $x$ by isolating it:
Subtract $y$ from both sides:
$\frac{x}{2} = 3 - y$
Multiply both sides by 2 to solve for $x$:
$x = 2(3 - y) = 6 - 2y$

Step 2: Substitute $x = 6 - 2y$ into the second equation $x + 2y = 6$:
$(6 - 2y) + 2y = 6$
Simplify:
$6 = 6$

Step 3: The equation $6 = 6$ is always true, indicating there is no contradiction and hence infinitely many solutions when both conditions arise from manipulating consistent equations.

Step 4: Since manipulating these equations leads us to an identity, they are dependent; both equations are forms of the same linear equation $\frac{x}{2} + y = 3$. Each point on this line satisfies both equations, confirming infinite solutions.

Therefore, the solution to the system of equations is infinite solutions.

To solve this system of linear equations, we'll employ the substitution method. The equations given are:

$\frac{x+3y}{2} = 0$

$x + y = 4$

Step 1: Solve the first equation for $x$.

The equation $\frac{x+3y}{2} = 0$ can be simplified:

Multiply both sides by 2 to eliminate the fraction:

$x + 3y = 0$

Solving for $x$, we get:

$x = -3y$

Step 2: Substitute this expression for $x$ into the second equation.

Substitute $x = -3y$ into $x + y = 4$:

$-3y + y = 4$

This simplifies to:

$-2y = 4$

Step 3: Solve for $y$.

Divide both sides by -2 to find $y$:

$y = -2$

Step 4: Substitute $y = -2$ back into the expression for $x$.

Using $x = -3y$:

$x = -3(-2) = 6$

Thus, the solution to the system of equations is $x = 6$ and $y = -2$.

Therefore, the solution to the problem is $x = 6, y = -2$.

To solve the system of equations, we will apply the substitution method:

We start with the given equations:

$4x - \frac{7}{y} = -3$ (Equation 1)

$5x + \frac{2}{y} = 7$ (Equation 2)

Let's start by solving Equation 1 for $\frac{1}{y}$:

$4x - \frac{7}{y} = -3$

Rearrange to isolate $\frac{1}{y}$:

$-\frac{7}{y} = -3 - 4x$

Multiply through by -1 to simplify:

$\frac{7}{y} = 4x + 3$

$\frac{1}{y} = \frac{4x + 3}{7}$ (Equation 3)

Now, substitute Equation 3 into Equation 2:

$5x + \frac{2}{y} = 7$

Replace $\frac{1}{y}$ from Equation 3:

$5x + 2 \left(\frac{4x + 3}{7}\right) = 7$

Multiply both sides of the equation by 7 to eliminate the fraction:

$35x + 2(4x + 3) = 49$

Expand and simplify:

$35x + 8x + 6 = 49$

$43x + 6 = 49$

Subtract 6 from both sides:

$43x = 43$

Divide by 43:

$x = 1$

Substitute $x = 1$ back into Equation 3 to solve for $y$:

$\frac{1}{y} = \frac{4(1) + 3}{7}$

$\frac{1}{y} = \frac{4 + 3}{7}$

$\frac{1}{y} = \frac{7}{7}$

$\frac{1}{y} = 1$

Therefore, $y = 1$

The solution to the system of equations is $x = 1, y = 1$.

To solve this system of equations, we'll use the substitution method.

Here are the steps we will take:

• Step 1: Solve the second equation $x + 3y = 6$ for $x$.
• Step 2: Substitute the expression for $x$ in the first equation $\frac{2x}{3} - y = 3$.
• Step 3: Solve the resulting equation for $y$.
• Step 4: Substitute back to find $x$.

Let's begin:

Step 1: Solve $x + 3y = 6$ for $x$:
$x = 6 - 3y$

Step 2: Substitute $x = 6 - 3y$ into the first equation $\frac{2x}{3} - y = 3$:
$\frac{2(6 - 3y)}{3} - y = 3$

Simplify the expression:
$\frac{12 - 6y}{3} - y = 3$

This simplifies to:
$4 - 2y - y = 3$

Combine like terms:
$4 - 3y = 3$

Isolate $y$:
$-3y = 3 - 4$
$-3y = -1$
$y = \frac{-1}{-3}$
$y = \frac{1}{3}$

Step 3: With $y = \frac{1}{3}$, substitute back into $x = 6 - 3y$:
$x = 6 - 3\left(\frac{1}{3}\right)$
$x = 6 - 1$
$x = 5$

Therefore, the solution to this system of equations is $\mathbf{x = 5, y = \frac{1}{3}}$.

Referring to the choice list, the correct choice is Choice 3: $x = 5, y = \frac{1}{3}$.

To solve this system of linear equations, follow these steps:

• Step 1: Examine and rearrange the first equation $\frac{3}{x} + \frac{3}{y} = 2$.
• Step 2: Express one variable in terms of the other. We can isolate $\frac{3}{x}$ to get $\frac{3}{x} = 2 - \frac{3}{y}$.
• Step 3: Substitute $\frac{3}{x} = 2 - \frac{3}{y}$ into the second equation $\frac{9}{x} - \frac{4}{y} = -7$.
• Step 4: Replace $\frac{9}{x}$ with $3\left(2 - \frac{3}{y}\right)$ leading to $3\left(2 - \frac{3}{y}\right) - \frac{4}{y} = -7$.
• Step 5: Simplify the equation: Distribute $3$ into $2 - \frac{3}{y}$ to obtain $(6 - \frac{9}{y}) - \frac{4}{y} = -7$.
• Step 6: Combine the terms to get $6 - \frac{13}{y} = -7$.
• Step 7: Isolate $\frac{13}{y}$: $\frac{13}{y} = 13$.
• Step 8: Solve for $y$: Multiply both sides by $y$, leading to $13 = 13y$, hence $y = 1$.
• Step 9: Substitute $y = 1$ back into $\frac{3}{x} = 2 - \frac{3}{y}$, resulting in $\frac{3}{x} = 2 - 3$, or $\frac{3}{x} = -1$.
• Step 10: Solve for $x$: Multiply both sides by $x$, leading to $3 = -x$, thus $x = -3$.

Therefore, the solution to the system of equations is $\boldsymbol{x = -3}$ and $\boldsymbol{y = 1}$.

The correct choice among the given answer choices is the third option: $\boldsymbol{x = -3, y = 1}$.

We begin by examining the given system of equations:

$\frac{3}{x} + \frac{1}{y} = 4$ --- (1)

$\frac{5}{x} - \frac{1}{y} = 4$ --- (2)

Let's eliminate $\frac{1}{y}$ by adding equations (1) and (2):

$\left(\frac{3}{x} + \frac{1}{y}\right) + \left(\frac{5}{x} - \frac{1}{y}\right) = 4 + 4$

$\frac{3}{x} + \frac{5}{x} = 8$

$\frac{8}{x} = 8$

Solving for $x$, we have:

$x = 1$

Now, substitute $x = 1$ back into equation (1):

$\frac{3}{1} + \frac{1}{y} = 4$

$3 + \frac{1}{y} = 4$

$\frac{1}{y} = 1$

Solving for $y$, we obtain:

$y = 1$

Thus, the solution to the system of equations is $x = 1$ and $y = 1$.

The correct answer choice is: $x=1,y=1$.