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I guess you mean simultaneous equations. There are three methods of solving these equations: graphing if the variables are x and y (though this is only an approximate method and should be avoided unless used to verify your answers, or if it's specified by the teacher), substitution (which should be used if one of the equations has one of the variables with a coefficient of 1), or the addition method (also known as elimination, as you try to eliminate one of the variables in order to find the value of the other).

Since I can't demonstrate the graphing method, I'll write out the other two using the equations:

2x - 5y = 1

3x + 5y = 14

SUBSTITUTION:

Here, you seek to substitute an alternate value of x or y into the equation from which you DID NOT extract the alternate value. Sounds complicated because I'm not very good at explaining. But I'll show it instead:

I want to eliminate that 2 that is in front of the x in the first equation, so that I can find what x equals to in terms of y, and then substitute that value into the second equation.

2x - 5y = 1, I'll divide the entire equation by 2:

x - 5y/2 = 1/2. I'll then rearrange the equation:

x = 1/2 + 5y/2, yay! Now we sub it into equation 2; replace the value of x with the value we got here.

3x + 5y = 14

3(1/2 + 5y/2) + 5y = 14, expand the brackets...

3/2 + 15y/2 + 5y = 14, multiply the entire equation by 2 to cancel out the denominator:

3 + 15y + 10y = 28, and collect the like terms:

25y = 25. Therefore:

y = 1

But we're not done yet! Now we have to sub this value of y into one of the equations, to find out what x is. Let's choose equation 1:

2x - 5y = 1

2x - 5 = 1

2x = 6

x = 3

And now we have our two answers: x = 3, y = 1.

ELIMINATION:

For this particular equation, elimination is a much more sensible method.

I want to make the coefficients of one of the variables equal in both equations, so that I can add the two equations together (or subtract one from the other) and eliminate that set of variables. In the equations I've provided, hell yeah! The coefficients of y are already equal. Or well, almost. One is +5 and the other is -5, meaning that I can add the two equations together to cancel out y. (Because -5+5=0, of course.)

(2x - 5y = 1) + (3x + 5y = 14), add all the like terms together to get:

2x + 3x = 1 + 14

5x = 15,

x = 3. Now I know this is correct because of the substitution method I did before.

Sub this value of x into one of the two equations like in the previous method, to get y = 1.

GRAPHING:

I'll just quickly summarise this; graphing can only be used if the two variables are x and y, because those are the only two variables in the equations of lines on a linear plane. So if you have a question with a, b, c, d and so on, just forget about this method entirely.

Otherwise, you can use this method to graph the two lines, and any points of intersections represent the values for x and y that solve both equations, in this case the two lines intersect at one point (3,1).

Unless your graphing skills are incredibly accurate, this method isn't very good for finding the answers to simultaneous equations. But otherwise, you can use this method to check if your answers are correct. For example, if you got an answer like x=1 and y=-6 for these two equations, you can then check the graph and know that something is incredibly wrong!

Hope this helps.

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13y ago
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11y ago

1k+4=5k-3

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