
Systems
of Linear Equations: Sections: Definitions, Solving by graphing, Substitition, Elimination/addition, Gaussian elimination. The addition method of solving systems of equations is also called the method of elimination. This method is similar to the method you probably learned for solving simple equations. If you had the equation "x + 6 = 11", you would write "–6" under either side of the equation, and then you'd "add down" to get "x = 5" as the solution. x + 6
= 11 You'll do something similar with the addition method.
2x + y = 9 Note that, if I add down, the y's will cancel out. So I'll draw an "equals" bar under the system, and add down: 2x + y = 9 Now I can divide through to solve for x = 5, and then backsolve, using either of the original equations, to find the value of y. The first equation has smaller numbers, so I'll backsolve in that one: 2(5) + y = 9 Then the solution is (x, y) = (5, –1). It doesn't matter which equation you use for the backsolving; you'll get the same answer either way. If I'd used the second equation, I'd have gotten: 3(5) – y = 16 ...which is the same result as before.
x – 2y = –9 Note that the xterms would cancel out if only they'd had opposite signs. I can create this cancellation by multiplying either one of the equations by –1, and then adding down as usual. It doesn't matter which equation I choose, as long as I am careful to multiply the –1 through the entire equation. (That means both sides of the "equals" sign!) I'll multiply the second equation. The "–1R_{2}" notation over the arrow indicates that I multiplied row 2 by –1. Now I can solve the equation "–5y = –25" to get y = 5. Backsolving in the first equation, I get: x –
2(5) = –9 Then the solution is (x, y) = (1, 5). A very common temptation is to write the solution in the form "(first number I found, second number I found)". Sometimes, though, as in this case, you find the yvalue first and then the xvalue second, and of course in points the xvalue comes first. So just be careful to write the coordinates for your solutions correctly. Copyright © Elizabeth Stapel 20032011 All Rights Reserved
2x – y = 9 Nothing cancels here, but I can multiply to create a cancellation. I can multiply the first equation by 4, and this will set up the yterms to cancel. Solving this, I get that x = 2. I'll use the first equation for backsolving, because the coefficients are smaller. 2(2) – y = 9 The solution is (x, y) = (2, –5).
4x – 3y = 25 Hmm... nothing cancels. But I can multiply to create a cancellation. In this case, neither variable is the obvious choice for cancellation. I can multiply to convert the xterms to 12x's or the yterms to 24y's. Since I'm lazy and 12 is smaller than 24, I'll multiply to cancel the xterms. (I would get the same answer in the end if I set up the yterms to cancel. It's not that how I'm doing it is "the right way"; it was just my choice. You could make a different choice, and that would be just as correct.) I will multiply the first row by 3 and the second row by 4; then I'll add down and solve.
Solving, I get that y = 5. Neither equation looks particularly better than the other for backsolving, so I'll flip a coin and use the first equation. 4x –
3(5) = 25 Remembering to put the xcoordinate first in the solution, I get: (x, y) = (10, 5) Usually when you are solving "by addition", you will need to create the cancellation. Warning: The most common mistake is to forget to multiply all the way through the equation, multiplying on both sides of the "equals" sign. Be careful of this.
12x – 13y = 2 I think I'll multiply the second equation by 2; this will at least get rid of the decimal place. Oops! This result isn't true! So this is an inconsistent system (two parallel lines) with no solution (with no intersection point). no solution
12x – 3y = 6 I think it'll be simplest to cancel off the yterms, so I'll multiply the second row by –3. Well, yes, but...? I already knew that zero equals zero. So this is a dependent system, and, solving for "y =", the solution is: y = 4x – 2 (Your text may format the answer as "(s, 4s – 2)", or something like that.) Remember the difference: a nonsense answer (like "0 = –2" in the previous problem) means an inconsistent system with no solution; a useless answer (like "0 = 0" above) means a dependent system where the whole line is the solution. Some books use only "x" and "y" for their variables, but many use additional variables. When you write the solution for an x,ypoint, you know that the xcoordinate goes first and the ycoordinate goes second. When you are dealing with other variables, assume (unless explicitly told otherwise) that those variables are written in alphabetical order. For instance, if the variables in a given system are a and b, the solution point would be (a, b); it would not be (b, a). Unless otherwise specified, the variables are written in alphabetical order. << Previous Top  1  2  3  4  5  6  7  Return to Index Next >>


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