Graphing Systems Of Equations: Estimate The Solution

Hey there, math enthusiasts! Today, we're diving into the visual world of algebra to estimate the solution to a system of equations by graphing. This method is super handy for getting a good idea of where the answer lies, especially when dealing with equations that might be a bit tricky to solve algebraically. We'll be using the following system:

$\begin{array}{l} 3 x+5 y=14 \\ 6 x-4 y=9 \end{array}$

And we'll be looking at a few potential solutions: A. $(\frac{7}{3},-\frac{7}{2})$, B. $(\frac{4}{3}, \frac{5}{2})$, and C. $(\frac{5}{2}, Discussion category : mathematics

Understanding Systems of Equations and Graphing

A system of equations is simply a set of two or more equations that share the same variables. When we talk about solving a system of equations, we're looking for the values of those variables that make all the equations in the system true simultaneously. In the context of graphing, each linear equation represents a straight line on a coordinate plane. The solution to a system of linear equations is the point (or points) where the lines intersect. That intersection point is the only point that lies on both lines, meaning it satisfies both equations.

Graphing is a powerful tool for visualizing these relationships. By plotting each equation on the same graph, we can see where they cross. This intersection point gives us an estimate of the solution. Why an estimate? Because reading a graph perfectly can be challenging, especially if the intersection point has fractional coordinates. However, graphing provides a fantastic visual check and can often pinpoint the solution with surprising accuracy, or at least narrow down the possibilities significantly.

To graph a linear equation, we usually need to convert it into slope-intercept form, which is $y = mx + b$. Here, '$m$' represents the slope of the line, and '$b$' is the y-intercept (the point where the line crosses the y-axis). Once we have the equations in this form, we can easily plot them. First, we plot the y-intercept. Then, using the slope (which tells us how much '$y$' changes for every one unit change in '$x$'), we can find other points on the line. Connecting these points forms our line.

Let's take our first equation: $3x + 5y = 14$. To get it into slope-intercept form, we need to isolate '$y$'.

1. Subtract $3x$ from both sides: $5y = -3x + 14$
2. Divide both sides by 5: $y = -\frac{3}{5}x + \frac{14}{5}$

So, the y-intercept for this line is $\frac{14}{5}$ (which is 2.8), and the slope is $-\frac{3}{5}$. This means for every 5 units we move to the right on the graph, we move 3 units down.

Now, let's do the same for our second equation: $6x - 4y = 9$.

1. Subtract $6x$ from both sides: $-4y = -6x + 9$
2. Divide both sides by -4: $y = \frac{-6}{-4}x + \frac{9}{-4}$
3. Simplify: $y = \frac{3}{2}x - \frac{9}{4}$

For this line, the y-intercept is $-\frac{9}{4}$ (which is -2.25), and the slope is $\frac{3}{2}$. This means for every 2 units we move to the right, we move 3 units up.

With both equations in slope-intercept form, we can now prepare to plot them. The y-intercepts are at 2.8 and -2.25, and the slopes are $-\frac{3}{5}$ and $\frac{3}{2}$. Visually comparing these, we can already see they are quite different, suggesting an intersection point that isn't immediately obvious. The negative slope of the first line means it goes downwards from left to right, while the positive slope of the second line means it goes upwards. This guarantees they will intersect at exactly one point. The challenge is to accurately sketch these lines and find that crossing point.

Plotting the Lines and Finding the Intersection

Now that we've converted our equations into slope-intercept form, $y = mx + b$, it's time to bring out the graph paper (or our digital graphing tool!) and plot these two lines. Remember, the solution to the system is the point where these two lines intersect.

Our first equation, $y = -\frac{3}{5}x + \frac{14}{5}$, has a y-intercept of $\frac{14}{5}$ (or 2.8) and a slope of $-\frac{3}{5}$.

• Start by plotting the y-intercept: Find 2.8 on the y-axis and mark that point. It's a bit above 2 and below 3.
• Use the slope to find another point: The slope is $-\frac{3}{5}$. This means 'rise' over 'run' is -3 over 5. From your y-intercept, go 5 units to the right (run = +5) and 3 units down (rise = -3). Plot this new point.
• Draw the line: Connect the y-intercept and the point you just found with a straight line. Extend it in both directions.

Our second equation, $y = \frac{3}{2}x - \frac{9}{4}$, has a y-intercept of $-\frac{9}{4}$ (or -2.25) and a slope of $\frac{3}{2}$.

• Plot the y-intercept: Find -2.25 on the y-axis. This point is between -2 and -3.
• Use the slope: The slope is $\frac{3}{2}$. From your y-intercept, go 2 units to the right (run = +2) and 3 units up (rise = +3). Plot this new point.
• Draw the line: Connect the y-intercept and this new point with a straight line. Extend it.

As you draw these lines, pay close attention to where they cross. The goal is to visually estimate the coordinates $(x, y)$ of this intersection point. Often, graphing can be imprecise, especially with fractional intercepts and slopes. You might find that the intersection doesn't fall exactly on integer grid lines, making estimation crucial.

Let's consider the options we've been given:

A. $(\frac{7}{3},-\frac{7}{2})$ which is approximately $(2.33, -3.5)$ B. $(\frac{4}{3}, \frac{5}{2})$ which is approximately $(1.33, 2.5)$ C. $(\frac{5}{2}, Discussion category : mathematics

Self-correction: Option C is incomplete. For the purpose of this exercise, we'll assume we are only evaluating options A and B as potential solutions to graph against.

If you were to sketch these lines carefully, you'd notice:

• The first line ($y = -\frac{3}{5}x + \frac{14}{5}$) passes through points like $(0, 2.8)$, $(5, -0.2)$, $(-5, 5.6)$.
• The second line ($y = \frac{3}{2}x - \frac{9}{4}$) passes through points like $(0, -2.25)$, $(2, 0.75)$, $(-2, -5.25)$.

When you plot these, you'll see the intersection point occurs where '$x$' is a positive value and '$y$' is also a positive value, but perhaps closer to zero for '$x$' and higher for '$y$'. The first line has a negative slope and a positive y-intercept, while the second has a positive slope and a negative y-intercept. This suggests they will intersect in the first quadrant (both x and y positive) or possibly the fourth quadrant (x positive, y negative) depending on the steepness of the slopes and the values of the intercepts.

Let's re-examine our options based on our visual estimation. Option A, $(2.33, -3.5)$, has a negative '$y$' value. Option B, $(1.33, 2.5)$, has both '$x$' and '$y$' values positive. Looking at the graph, the intersection point appears to be in the region where both '$x$' and '$y$' are positive. The first line starts high on the y-axis and goes down, while the second starts low on the y-axis and goes up. They will likely cross in the upper-right quadrant.

By carefully plotting these lines, you'd visually determine that the intersection point is much closer to option B than option A. Option A's negative '$y$' value doesn't align with where the lines appear to cross. Option B's positive '$x$' and '$y$' values seem more plausible.

Verifying the Estimated Solution

While graphing gives us a great estimate of the solution, the best way to be sure is to verify our suspected answer by plugging the coordinates back into the original equations. This is a crucial step because reading a graph perfectly can be difficult, and our visual estimation might be slightly off. We want to find the point that makes both equations true.

Let's test option B: $(\frac{4}{3}, \frac{5}{2})$.

Equation 1: $3x + 5y = 14$ Substitute $x = \frac{4}{3}$ and $y = \frac{5}{2}$: $3(\frac{4}{3}) + 5(\frac{5}{2}) = ? 14$ $4 + \frac{25}{2} = ? 14$ $4 + 12.5 = ? 14$ $16.5 \neq 14$

Uh oh! It looks like option B doesn't perfectly satisfy the first equation. This tells us that either our graphical estimation was a bit off, or option B isn't the exact solution, even if it looked close on the graph. Let's re-evaluate our graphical estimation and perhaps consider the possibility that the intersection isn't exactly one of the provided choices but that we should pick the closest estimate. Or, let's check the other options just in case.

Let's test option A: $(\frac{7}{3},-\frac{7}{2})$.

Equation 1: $3x + 5y = 14$ Substitute $x = \frac{7}{3}$ and $y = -\frac{7}{2}$: $3(\frac{7}{3}) + 5(-\frac{7}{2}) = ? 14$ $7 - \frac{35}{2} = ? 14$ $7 - 17.5 = ? 14$ $-10.5 \neq 14$

Neither option A nor B appears to be the exact solution when plugged back into the first equation. This is a good lesson: graphing provides an estimate. If the options given are meant to be exact solutions, then our graphing was either not precise enough, or there might be a slight issue with the options provided in relation to the exact graphical solution. However, the question asks us to estimate the solution by graphing. Let's reconsider the graph.

Looking at the slopes and intercepts again: Line 1: $y = -\frac{3}{5}x + \frac{14}{5}$ (y-int = 2.8, slope = -0.6) Line 2: $y = \frac{3}{2}x - \frac{9}{4}$ (y-int = -2.25, slope = 1.5)

We expect the intersection to have $x > 0$ and $y > 0$ because Line 1 starts high and goes down, and Line 2 starts low and goes up, and they must cross somewhere in between.

Let's see if we can find the exact solution algebraically to compare. We can use substitution or elimination.

Using elimination: Multiply the first equation by 2: $6x + 10y = 28$ Keep the second equation: $6x - 4y = 9$ Subtract the second equation from the first: $(6x + 10y) - (6x - 4y) = 28 - 9$ $14y = 19$ $y = \frac{19}{14}$

Now substitute $y = \frac{19}{14}$ into the first original equation: $3x + 5(\frac{19}{14}) = 14$ $3x + \frac{95}{14} = 14$ $3x = 14 - \frac{95}{14}$ $3x = \frac{196}{14} - \frac{95}{14}$ $3x = \frac{101}{14}$ $x = \frac{101}{42}$

The exact solution is $(\frac{101}{42}, \frac{19}{14})$.

Let's convert this to decimals to compare with our options: $x = \frac{101}{42} \approx 2.40$ $y = \frac{19}{14} \approx 1.36$

So the exact solution is approximately $(2.40, 1.36)$.

Now let's look at our options again and their decimal approximations:

• A. $(\frac{7}{3},-\frac{7}{2}) \approx (2.33, -3.5)$
• B. $(\frac{4}{3}, \frac{5}{2}) \approx (1.33, 2.5)$

Comparing our estimated solution $(2.40, 1.36)$ to the options:

• Option A $(2.33, -3.5)$: The x-value is close, but the y-value is very different (1.36 vs -3.5).
• Option B $(1.33, 2.5)$: Both x and y values are quite far off from the exact solution (2.40 vs 1.33 for x, and 1.36 vs 2.5 for y).

This discrepancy suggests that either the provided options are not close estimates, or the intended graphical estimation method leads to a different conclusion. When estimating by graphing, we rely on visual accuracy. Let's imagine plotting the lines carefully.

Line 1: $y = -0.6x + 2.8$ Line 2: $y = 1.5x - 2.25$

If we plot these, Line 1 starts at (0, 2.8) and goes down. Line 2 starts at (0, -2.25) and goes up. The intersection must occur where $x > 0$. Let's test the x-values of our options.

Option A has $x = 7/3 \approx 2.33$. Option B has $x = 4/3 \approx 1.33$.

If $x = 2.33$ (Option A), Line 1 y-value: $-0.6(2.33) + 2.8 \approx -1.4 + 2.8 = 1.4$. Line 2 y-value: $1.5(2.33) - 2.25 \approx 3.5 - 2.25 = 1.25$. These y-values (1.4 and 1.25) are close to each other, and the actual y-value of the intersection is $1.36$. This suggests that $x=7/3$ is a good estimate for the x-coordinate.

If $x = 1.33$ (Option B), Line 1 y-value: $-0.6(1.33) + 2.8 \approx -0.8 + 2.8 = 2.0$. Line 2 y-value: $1.5(1.33) - 2.25 \approx 2.0 - 2.25 = -0.25$. These y-values (2.0 and -0.25) are very different, indicating $x=1.33$ is not a good estimate for the x-coordinate.

Therefore, based on estimating the x-coordinate by graphing, $x = \frac{7}{3}$ seems to be the better estimate.

Now let's look at the y-coordinates for $x = \frac{7}{3}$. We calculated the exact y-value as $1.36$. Option A gives $y = -7/2 = -3.5$. Option B gives $y = 5/2 = 2.5$.

Neither of these y-values is close to $1.36$. This is quite confusing! Let's re-read the problem. It asks to estimate the solution by graphing. This implies we should visually find the intersection. The exact solution is $(\frac{101}{42}, \frac{19}{14})$.

Let's check option C, assuming it was meant to be a valid point. If we assume there was a typo and option C was something like $(\frac{101}{42}, \frac{19}{14})$ or a close approximation, that would be the answer. Since we don't have a complete option C, we must choose between A and B based on our graphical estimation.

When plotting the lines, we see Line 1 ($y = -0.6x + 2.8$) starts at (0, 2.8) and Line 2 ($y = 1.5x - 2.25$) starts at (0, -2.25). The slope of Line 2 is much steeper than Line 1. This means Line 2 will rise much faster than Line 1 falls. The intersection point will likely have an x-value that is not too large, and a y-value that is positive but not excessively high.

Let's reconsider our test of option A: $(\frac{7}{3}, -\frac{7}{2}) \approx (2.33, -3.5)$. The x-value is plausible, but the y-value is negative, which doesn't seem right for the intersection based on the intercepts.

Let's reconsider our test of option B: $(\frac{4}{3}, \frac{5}{2}) \approx (1.33, 2.5)$. The x-value is smaller, and the y-value is positive. Let's check how close this point is to satisfying the equations:

For $3x+5y=14$ with $(1.33, 2.5)$: $3(1.33) + 5(2.5) = 3.99 + 12.5 = 16.49$ (Close to 14, but a bit high) For $6x-4y=9$ with $(1.33, 2.5)$: $6(1.33) - 4(2.5) = 7.98 - 10 = -2.02$ (Not close to 9)

There seems to be a significant issue with the provided options in relation to the problem. However, if forced to choose the best graphical estimate between A and B, and assuming the intention was for one of them to be a reasonable approximation, we must rely on the visual representation.

Visually, the intersection point appears to be in the first quadrant (x>0, y>0). Option A has a negative y-value, making it unlikely. Option B has positive x and y values. Even though our algebraic check showed B wasn't a perfect fit, graphical estimation can sometimes be misleading if not done with extreme precision.

Let's focus on the x-values. We found the exact x is $\approx 2.40$. Option A's x is $\approx 2.33$, which is very close. Option B's x is $\approx 1.33$, which is not close.

Now let's focus on the y-values. We found the exact y is $\approx 1.36$. Option A's y is $-3.5$. Option B's y is $2.5$.

Option A's x-value is the best estimate. Option B's y-value is closer to the correct y-value than Option A's y-value, but Option A's x-value is much closer. When graphing, the intersection point is what matters. The point $(2.33, -3.5)$ (Option A) is in the fourth quadrant. The point $(1.33, 2.5)$ (Option B) is in the first quadrant. Based on the intercepts (positive for line 1, negative for line 2) and slopes (negative for line 1, positive for line 2), the intersection should be in the first quadrant.

Given this, Option B, despite its inaccuracies in algebraic verification, is the more likely candidate for a graphical estimate because it falls in the expected quadrant. The error in algebraic verification suggests that the options might not be precise, or the intended answer is based purely on visual approximation in the correct quadrant.

For a truly accurate graphical estimation, one would need to plot the lines very precisely. However, based on the general positions and slopes, the intersection appears to be in the first quadrant, favoring Option B as the estimated solution, even if it doesn't verify perfectly.

Conclusion: The Power of Visual Estimation

Estimating the solution to a system of equations by graphing is a valuable skill that bridges the gap between algebraic manipulation and geometric understanding. We've seen that while graphing might not always yield an exact algebraic answer, it provides a crucial visual checkpoint. By converting our equations into slope-intercept form, plotting the lines, and observing their intersection, we can pinpoint an approximate solution.

In this case, the system $3x + 5y = 14$ and $6x - 4y = 9$ led us through the process of plotting. We transformed them into $y = -\frac{3}{5}x + \frac{14}{5}$ and $y = \frac{3}{2}x - \frac{9}{4}$. The visual intersection of these lines appears to be in the first quadrant, where both '$x$' and '$y$' are positive.

Considering the provided options, A. $(\frac{7}{3},-\frac{7}{2})$ and B. $(\frac{4}{3}, \frac{5}{2})$, our analysis of the graphical placement strongly suggests that option B is the better estimate. Although algebraic verification revealed that neither option is the exact solution, the quadrant of intersection derived from the graph is a powerful indicator. Option A's negative '$y$' value places it in the wrong quadrant, whereas Option B correctly suggests a point in the first quadrant. This highlights how graphical estimation can help eliminate incorrect answers and guide us toward the most plausible solution.

Always remember that graphing is an estimation technique. For precise answers, algebraic methods like substitution or elimination are necessary. However, for a quick understanding of where the solution lies, or to check the reasonableness of an algebraic solution, graphing is indispensable.

If you're looking to hone your skills in solving systems of equations, I recommend checking out resources like Khan Academy. They offer comprehensive guides and practice exercises on graphing linear equations and solving systems of equations, which can greatly enhance your understanding. For more in-depth mathematical exploration, Wolfram Alpha is an incredible tool for verifying solutions and exploring complex mathematical problems.