Systems Of Equations: Finding No Solution

Let's dive into the world of systems of equations and figure out which ones don't have a solution. It's like searching for a treasure that isn't there – a fun mathematical quest! We will analyze each option step by step to determine if they represent inconsistent systems.

Understanding Systems of Equations

Before we start, let's recap what a system of equations is. A system of equations is a set of two or more equations containing the same variables. The solution to a system of equations is the set of values for the variables that satisfy all equations simultaneously. Graphically, the solution represents the point(s) where the lines intersect. When a system has no solution, it means the lines are parallel and never intersect. This typically occurs when the equations represent lines with the same slope but different y-intercepts. Identifying such systems involves manipulating the equations into slope-intercept form (y = mx + b) or observing the ratios of coefficients.

Now, let’s break down each option to see which systems of equations lead to this no-solution scenario.

A. $x - y = 1, -3x + 3y = 3$

Let's analyze the first system: $x - y = 1$ and $-3x + 3y = 3$. To determine if this system has no solution, one solution, or infinitely many solutions, we can manipulate the equations to see if they are consistent or inconsistent. First, let's solve the first equation for $y$: $y = x - 1$. Now, let's look at the second equation, $-3x + 3y = 3$. We can divide the entire equation by 3 to simplify it: $-x + y = 1$. Solving for $y$, we get $y = x + 1$. Now we have two equations in slope-intercept form: $y = x - 1$ and $y = x + 1$. Notice that both lines have the same slope (1) but different y-intercepts (-1 and 1, respectively). This means the lines are parallel and will never intersect. Therefore, this system has no solution. This occurs because the equations are inconsistent; they describe parallel lines.

In summary, when we manipulate the given equations, we find that they represent parallel lines. Parallel lines never intersect, which means there is no point (x, y) that satisfies both equations simultaneously. Hence, the system has no solution. Systems with no solutions are called inconsistent systems. This often happens when the coefficients of x and y are proportional, but the constants are not, indicating parallel lines. Therefore, option A is a valid system with no solutions.

B. $2y = 3 - 4x, x + y = 0$

Next, let's examine the system $2y = 3 - 4x$ and $x + y = 0$. We'll convert both equations into slope-intercept form ($y = mx + b$) to compare their slopes and y-intercepts. First, let's rewrite the first equation: $2y = 3 - 4x$. Divide both sides by 2 to isolate $y$: $y = \frac{3}{2} - 2x$, which can be written as $y = -2x + \frac{3}{2}$. Now, let's rewrite the second equation: $x + y = 0$. Subtract $x$ from both sides to solve for $y$: $y = -x$. Now we have two equations in slope-intercept form: $y = -2x + \frac{3}{2}$ and $y = -x$. The slopes of these lines are -2 and -1, respectively. Since the slopes are different, the lines are not parallel and will intersect at a single point. Therefore, this system has one solution. The lines are neither parallel nor the same line, so they must intersect at exactly one point. Graphically, you would see two distinct lines crossing each other.

In conclusion, the two lines in this system have different slopes, ensuring they intersect at exactly one point. Thus, the system has a unique solution and is not a system with no solution. Therefore, option B is not a valid choice.

C. $12x + 2 = 4y - 10, 2x + y = -11$

Now, let's consider the system $12x + 2 = 4y - 10$ and $2x + y = -11$. Again, we'll convert both equations into slope-intercept form to analyze them. First, let's rewrite the first equation: $12x + 2 = 4y - 10$. Add 10 to both sides: $12x + 12 = 4y$. Divide both sides by 4 to isolate $y$: $3x + 3 = y$, which can be written as $y = 3x + 3$. Now, let's rewrite the second equation: $2x + y = -11$. Subtract $2x$ from both sides to solve for $y$: $y = -2x - 11$. Now we have two equations in slope-intercept form: $y = 3x + 3$ and $y = -2x - 11$. The slopes of these lines are 3 and -2, respectively. Since the slopes are different, the lines are not parallel and will intersect at a single point. Therefore, this system has one solution. The equations represent two distinct lines that intersect at a unique point, giving a single solution.

In summary, because the slopes of the two lines are different, they are guaranteed to intersect at one point. Therefore, the system has one unique solution, and option C is not a system with no solutions.

D. $-3x - y = 5, 15x = 10 - 5y$

Let's analyze the system $-3x - y = 5$ and $15x = 10 - 5y$. We'll transform both equations into slope-intercept form. First, let's rewrite the first equation: $-3x - y = 5$. Add $3x$ to both sides: $-y = 3x + 5$. Multiply both sides by -1 to solve for $y$: $y = -3x - 5$. Now, let's rewrite the second equation: $15x = 10 - 5y$. Add $5y$ to both sides: $15x + 5y = 10$. Subtract $15x$ from both sides: $5y = -15x + 10$. Divide both sides by 5 to isolate $y$: $y = -3x + 2$. Now we have two equations in slope-intercept form: $y = -3x - 5$ and $y = -3x + 2$. Notice that both lines have the same slope (-3) but different y-intercepts (-5 and 2, respectively). This means the lines are parallel and will never intersect. Therefore, this system has no solution. The two lines are parallel but have different y-intercepts, meaning they will never intersect.

In conclusion, the two lines are parallel, as they have the same slope but different y-intercepts. Consequently, there is no solution to the system of equations, and option D is a valid choice.

E. $x + y = 1, -4x + 2y = 7$

Finally, let's analyze the system $x + y = 1$ and $-4x + 2y = 7$. We'll convert both equations into slope-intercept form. First, let's rewrite the first equation: $x + y = 1$. Subtract $x$ from both sides to solve for $y$: $y = -x + 1$. Now, let's rewrite the second equation: $-4x + 2y = 7$. Add $4x$ to both sides: $2y = 4x + 7$. Divide both sides by 2 to isolate $y$: $y = 2x + \frac{7}{2}$. Now we have two equations in slope-intercept form: $y = -x + 1$ and $y = 2x + \frac{7}{2}$. The slopes of these lines are -1 and 2, respectively. Since the slopes are different, the lines are not parallel and will intersect at a single point. Therefore, this system has one solution. These lines will intersect at a single point because their slopes are distinct.

In conclusion, the slopes of these lines are distinct, which means they intersect at a single point. Thus, the system has one solution, and option E is not a valid choice.

Final Answer

Based on our analysis, the systems of equations that have no solution are:

• A. $x - y = 1, -3x + 3y = 3$
• D. $-3x - y = 5, 15x = 10 - 5y$

These systems represent parallel lines that never intersect, resulting in no common solution.

To further explore systems of equations, you can visit Khan Academy's section on systems of equations for more practice and explanations.