Solve graphically the system of linear equations:

x+y=0 −3x+4y=14

Draw Cartesian coordinate system.

Find two points on line x + y = 0

For example:

x = 5

x + y = 0

5 + y = 0

subtract 5 to both sides

y = - 5

x = - 3

x + y = 0

- 3 + y = 0

Add 3 to both sides

y = 3

Mark points ( 5 , - 5 ) ( - 3 , 3 ) and draw straight line.

Find two points on line 3 x + 4 y = 14

x = 0

3 x + 4 y = 14

3 ∙ 0 + 4 y = 14

0 + 4 y = 14

4 y = 14

y = 14 / 4 = 2 ∙ 7 / 2 ∙ 2 = 7 / 2 = 3.5

y = 0

3 x + 4 y = 14

3 x + 4 ∙ 0 = 14

3 x + 0 = 14

3 x = 14

x = 14 / 3 = 4.66666

Mark points ( 0 , 3.5 ) ( 4.66666 , 0 ) and draw straight line.

The x and y coordinates of the intersection will be the solution to the system of equations.

The solution is x = - 2 , y = 2

Oh, I see you want me to solve the system of linear equations using humor. Alright, let's give it a shot!

Well, the first equation is x + y = 0. So, we can see that x and y must have a cozy relationship and always add up to zero. They must be the ultimate example of mathematical teamwork!

Now, let's move on to the second equation: -3x + 4y = 14. Hmm, it seems like -3x is trying to drag x down while 4y is doing its best to lift y up. Talk about a love-hate relationship between the variables!

To solve these two equations graphically, we'll need to draw their respective lines. So, let's grab our drawing tools and get started!

After plotting the lines on the graph, we find that they intersect at a point. And that point, my friend, is the solution to this system of linear equations.

But hey, don't worry if you don't have the graph with you right now. If you need the actual numerical solution, I'm here to help you out. Just let me know!

To solve the system of linear equations graphically, first let's plot the lines representing each equation on a graph.

For the equation x + y = 0, rearrange it to y = -x.

For the equation -3x + 4y = 14, rearrange it to 4y = 3x + 14 and then y = (3/4)x + 7/2.

Now, plot the points on the graph:

For y = -x:
Let's choose a few values for x and calculate the corresponding y values:

When x = -3, y = 3
When x = 0, y = 0
When x = 3, y = -3

Plot these points on the graph and draw a line connecting them.

For y = (3/4)x + 7/2:
Choose a few values for x and calculate the corresponding y values:

When x = -4, y = 4
When x = 0, y = 7/2
When x = 4, y = 13/2

Plot these points on the graph and draw a line connecting them.

Now, the graph should have two lines intersecting at a single point. The coordinates of that point represent the solution to the system of linear equations.

In this case, the solution is (2, -2).

Therefore, the system of linear equations x + y = 0 and -3x + 4y = 14 is solved graphically by finding the intersection point (2, -2).

To solve the system of linear equations graphically, we need to plot the equations on a graph and find the intersection point, which represents the solution.

Let's start by solving the first equation, x + y = 0.

To plot this equation graphically, we can rewrite it in the form y = mx + b, where m is the slope and b is the y-intercept. The equation can be rearranged to y = -x.

To plot this line, we can assign arbitrary values to x and solve for y. For example, when x = 0, y = 0. When x = 1, y = -1. When x = -1, y = 1. These points can be plotted on the graph.

Now let's move on to the second equation, -3x + 4y = 14.

Similarly, we can rewrite this equation in slope-intercept form. First, isolate y by rearranging the equation: 4y = 3x + 14. Then, divide both sides by 4: y = (3/4)x + 14/4, which simplifies to y = (3/4)x + 7/2.

To plot this line, we can again assign arbitrary values to x and solve for y. For example, when x = 0, y = 7/2. When x = 4, y = 15/2. When x = -4, y = -1/2.

Now that we have both lines plotted on the graph, we can visually determine the intersection point. This point represents the solution to the system of equations.

Looking at the graph, we see that the two lines intersect at the point (-2, 2). Therefore, the solution to the system of linear equations is x = -2 and y = 2.

Note that this is just a graphical approximation. For a more accurate solution, you can also solve the system algebraically by using methods such as substitution or elimination. The graphical method is helpful for visualizing the solution and providing an initial estimate.

there are many fine online graphing sites.

desmos.com is a handy one.