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systems of equations

2x+y=-2 -1x+y=4

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Solving a System of Equations


Home » Systems of Equations System of Equations. So, what is a system of equations? This may be a new term for you if you are just beginning your study of Algebra.
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Systems of Linear Equations - Maths Resources


Systems of Linear Equations . A Linear Equation is an equation for a line. A System of Equations is when we have two or more equations working together.
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Systems of equations | Algebra basics | Math | Khan Academy


Solving a system of equations or inequalities in two variables by elimination, substitution, and graphing. Systems of equations: trolls, tolls (1 of 2) Systems of ...
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SYSTEMS OF EQUATIONS - S.O.S. Mathematics


A system of equations is a collection of two or more equations with a same set of unknowns. In solving a system of equations, we try to find values for each of the ...
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Solve Systems of Equations - Tutorial


Solve Systems of Equations - Tutorial. This is a tutorial on solving 2 by 2 systems of linear equations. Detailed solutions and explanations are provided.
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Linear systems of equations capstone (practice) | Khan Academy


Solve systems of equations with any number of solutions using any solution method.
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Systems of Linear Equations, Solutions examples, pictures and ...


Systems of linear equations and their solution, explained with pictures , examples and a cool interactive applet. Also, a look at the using substitution, graphing and ...
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Systems of Linear Equations: Definitions


A "system" of equations is a set or collection of equations that you deal with all together at once. Linear equations (ones that graph as straight ...
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Solving Systems of Equations Using Algebra Calculator - MathPapa


Learn how to use the Algebra Calculator to solve systems of equations. Example Problem Solve the following system of equations: x+y=7, x+2y=11
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Systems of Equations - SparkNotes


Systems of Equations . We have worked with two types of equations--equations with one variable and equations with two variables. In general, we could find a limited ...
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Suggested Questions And Answer :


2x - y = -7 4x+4y=-8

We appear to have two systems of equations, but the method for solving either is essentially the same. First we make the equations into the form y=. The first system: y=2x+7, 4y=-8-4x or y=-2-x.  The second system: y=3-2x, y=2-x. Now we draw the graphs. We can put them all on to the same graph, but we should label all the line graphs so as not to get them mixed up. To draw straight line graphs, we find the x and y intercepts and draw a line joining and passing through them. System 1: y=2x+7: y-int (x=0) is 7, x-int (y=0) is -3.5; y=-2-x: y-int is -2, x-int is -2; system 2: y=3-2x: y-int is 3, x-int is 2; y=2-x: y-int is 2, x-int is 2. For each system, draw a line through the intercepts for each equation. System 1: join 7 on the y-axis to -3.5 on the x axis and join -2 on y to -2 on x; system 2: join 3 on y to 2 on x and join 2 on y to 2 on x. Label the lines with their equations. You should find that in system 1 the lines cross at the solution to 2x+7=-2-x, 3x=-9, so x=-3 and y=1 (-2+3 or -6+7); and in system 2 3-2x=2-x, x=1 and y=1. The points where the lines cross are therefore (-3,1) for system 1 and (1,1) for system 2. These are the graphical representations of the solutions of the two systems of equations.
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How do I solve equations simultaneously?

In mathematics, simultaneous equations and systems of equations are finite sets of equations whose common solutions are looked for. The systems of equations are usually classified in the same way as the single equations, namely: System of linear equations System of polynomial equations System of ordinary differential equations System of partial differential equations
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Find the orthogonal canonical reduction of the quadratic form −x^2 +y^2 +z^2 −6xy−6xz+2yz. Also, find its principal axes, rank and signature of the quadratic form.

4.4 Systems of Equations - Three Variables Objective: Solve systems of equations with three variables using addi- tion/elimination. Solving systems of equations with 3 variables is very simila r to how we solve sys- tems with two variables. When we had two variables we reduced the system down to one with only one variable (by substitution or addition). With three variables we will reduce the system down to one with two variables (usua lly by addition), which we can then solve by either addition or substitution. To reduce from three variables down to two it is very importan t to keep the work organized. We will use addition with two equations to elimin ate one variable. This new equation we will call (A). Then we will use a different pair of equations and use addition to eliminate the same variable. This second new equation we will call (B). Once we have done this we will have two equation s (A) and (B) with the same two variables that we can solve using either met hod. This is shown in the following examples. Example 1. 3 x + 2 y − z = − 1 − 2 x − 2 y + 3 z = 5 We will eliminate y using two different pairs of equations 5 x + 2 y − z =
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Solve the system using the elimination method.

Solve the system using the elimination method. Solve the system using the elimination method.  -8x    - 4y     + 2z    =     -26 -2x    + 4y    + 2z    =     10 -6x    - 8y    - 5z    =     -41 It appears that the three equations are: 1)  -8x - 4y + 2z = -26 2)  -2x + 4y + 2z = 10 3)  -6x - 8y - 5z = -41 Subtract equation two from equation one, eliminating the z.   -8x - 4y + 2z = -26 -(-2x + 4y + 2z =  10) ---------------------------   -6x   -  8y        = -36 4)  -6x - 8y = -36 Multiply equation two by 5. 5 * (-2x + 4y + 2z) = 10 * 5 5)  -10x + 20y + 10z = 50 Multiply equation three by 2. 2 * (-6x - 8y - 5z) = -41 * 2 6) -12x - 16y - 10z = -82 Add equation six to equation five, again eliminating the z.    -10x + 20y + 10z = 50 +(-12x - 16y - 10z = -82) --------------------------------    -22x  + 4y           = -32 7)  -22x + 4y = -32 Multiply equation seven by 2. 2 * (-22x + 4y) = -32 * 2 8)  -44x + 8y = -64 Add equation eight to equation four, eliminating the y.      -6x -  8y = -36 +(-44x + 8y = -64) -----------------------    -50x         = -100 -50x = -100 x = 2  <<<<<<<<<<<<<<<<<<<< Substitute the value of x into equation four. -6x - 8y = -36 -6(2) - 8y = -36 -12 - 8y = -36 -8y = -36 + 12 -8y = -24 y = 3  <<<<<<<<<<<<<<<<<<<< Substitute both x and y into equation one. -8x - 4y + 2z = -26 -8(2) - 4(3) + 2z = -26 -16 - 12 + 2z = -26 -28 + 2z = -26 2z = -26 + 28 2z = 2 z = 1  <<<<<<<<<<<<<<<<<<<< Substitute all three values into equation two. -2x + 4y + 2z = 10 -2(2) + 4(3) + 2(1) = 10 -4 + 12 + 2 = 10 -4 + 14 = 10 10 = 10 Substitute all three values into equation three. -6x - 8y - 5z = -41 -6(2) - 8(3) - 5(1) = -41 -12 - 24 - 5 = -41 -36 - 5 = -41 -41 = -41 Everything checks. x = 2, y = 3, z = 1
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solve following system of equations

Problem: solve following system of equations solve the following system of equations: x-2y+z=6 , 2x+y-3z=-3, x-3y+3z=10 1) x - 2y + z = 6 2) 2x + y - 3z = -3 3) x - 3y + 3z = 10 Multiply equation 1 by 3. 3(x - 2y + z) = 6 * 3 4) 3x - 6y + 3z = 18 Add equation 2 to equation 4.   3x - 6y + 3z = 18 +(2x +  y - 3z = -3) -----------------------   5x - 5y      = 15 6) 5x - 5y = 15 Add equation 3 to equation 2.   2x +  y - 3z = -3 +( x - 3y + 3z = 10) ----------------------   3x - 2y      = 7 7) 3x - 2y = 7 Multiply equation 6 by 2. 2(5x - 5y) = 15 * 2 8)10x - 10y = 30 Multiply equation 7 by 5. 5(3x - 2y) = 7 * 5 9) 15x - 10y = 35 Subtract equation 9 from equation 8.   10x - 10y = 30 -(15x - 10y = 35) -------------------   -5x       = -5 -5x = -5 -5x/-5 = -5/-5 x = 1 Use equation 6 to solve for y. 5x - 5y = 15 5(1) - 5y = 15 5 - 5y = 15 5 - 5y - 5 = 15 - 5 -5y = 10 -5y/-5 = 10/-5 y = -2 Use equation 3 to solve for z. x - 3y + 3z = 10 1 - 3(-2) + 3z = 10 1 + 6 + 3z = 10 7 + 3z = 10 3z = 3 3z/3 = 3/3 z = 1 Answer: x = 1, y = -2, z = 1  
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how do you use brackets in simultaneous equations

One reason you might use brackets in simultaneous equations is for expressing the solution as an ordered set (usually pairs). If, for example, you had a 2-variable system of equations, x and y, you might express the result as the ordered pair (x,y) where x is replaced by the value found for x and y the value found for y. Sometimes a system of equations has more than one solution, so the different solutions would be represented as (x1,y1) and (x2,y2). This can happen when the system involves one or more quadratic equations. The brackets ensure that the values for x and y are not mixed up. Another reason for brackets is when using substitution to solve a system. Suppose there are two equations: ax+by=c and dx+ey=f. From the first equations we can write y=(c-ax)/b and substitute for y in the second equation: dx+e(c-ax)/b=f. That would be the first step in solving. The next step would be to expand the brackets and solve for x in terms of the constants a, b, c, d, e and f. Having found x, you find y by substituting the value of x in y=(c-ax)/b. Sometimes a question involving simultaneous equations comes in a form that uses brackets, so the first step is to expand the brackets, combine like terms, and then continue to solve the system.
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What can you conclude about the system of equations?

Consider the following system of equations.  What can you conclude about the system of equations?  y=3x-7  y-3x=5 All linear equations are of the form: y = mx + c, where m is the slope of the graph and c is the y-intercept. Our two equations, when rewritten, are, y = 3x - 7 y = 3x + 5 It can be seen that they have the same gradient, m = 3, and so they are parallel. The 1st line has a y-intercept of c = -7, while the second line has a y-intercept of c = 5. The equations represent two parallel lines a vertical distance of 12 units apart.
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Solving systems of equations by addition

The first step is to make sure the equations are setup such that either the x terms or the y terms are opposites of each other (here, +13y and -13y satisfy this condition, but in general you may need to multiply an entire equation by a certain number to make this happen).  Next, add one equation to the other like so: 4x + 13y = 16 2x - 13y = 8 -------------------- 6x  +  0y   = 24   Note, the coefficient on the y term is zero (a result of having the opposites set up).  This is important because we went from a system of two equations with two unknowns, to one equation with one unknown, namely, 6x = 24) Solve 6x=24 and get x=4.  Now plug x=4 into one of the original equations and find the y value. 2x - 13y = 8   ==>  2(4) - 13y = 8  ==> 8 - 13y = 8  ==>  -13y = 0  ==>  y=0 The solution to the system of equations is x=4, y=0. Check this solution by plugging both x=4 and y=0 into both equations, and verify you get true statements. 4x + 13y = 16   ==>   4(4) + 13(0)  = 16 + 0 = 16   True 2x - 13y = 8    ==>   2(4) - 13(0) = 8 - 0 = 8   True
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Find the exact solutions of each system of equations 4x+y^2=20 4x^2+y^2=100

Find the exact solutions of each system of equations 4x+y^2=20 4x^2+y^2=100 We need to find the exact solutions of each system of equations written as ordered pairs. 1) 4x+y^2=20 2) 4x^2+y^2=100 Subtract equation one from equation two, eliminating y^2.   4x^2    + y^2 = 100 -(     4x + y^2 =  20) ------------------------   4x^2 - 4x     = 80 4x^2 - 4x = 80 3) 4x^2 - 4x - 80 = 0 Solve equation three for x. Factor the left side. 4x^2 - 4x - 80 = 0 (4x - 20)(x + 4) = 0 One or both factors must equal zero. Factor one: (4x - 20) = 0 4x = 20 x = 5 Factor two: (x + 4) = 0 x = -4 Plug both of those values into equation one to solve for y. 4x + y^2 = 20 4(5) + y^2 = 20 20 + y^2 = 20 y^2 = 0 y = 0 (5, 0) 4x + y^2 = 20 4(-4) + y^2 = 20 -16 + y^2 = 20 y^2 = 20 + 16 y^2 = 36 y = ±6     (you can get 36 by squaring 6 and by squaring negative 6) (-4, -6) (-4, 6) Use equation two to verify the various values. (5, 0) 4x^2 + y^2 = 100 4(5^2) + 0^2 = 100 4(25) + 0 = 100 100 = 100 (-4, -6) 4x^2 + y^2 = 100 4(-4^2) + (-6^2) = 100 4(16) + (36) = 100 64 + 36 = 100 100 = 100 (-4, 6) 4x^2 + y^2 = 100 4(-4^2) + 6^2 = 100 4(16) + 36 = 100 64 + 36 = 100 100 = 100 All three pairs are valid: (5, 0), (-4, -6) and (-4, 6)
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The point (4, -2) is the solution to which of the following system of equations?

The point (4, -2) is the solution to which of the following system of equations? help me solve this someone? The point  (4, -2) is the solution to which of the following system of equations? A) Y =-X +2     Y = 4X -2 B) Y = -X + 2      Y= -2 +4 C) Y = X -2       Y = X-6 D) Y =2X -10       Y = X -6 All you have to do is substitute the values of x and y, given as the point, into the equations and find out which set works. A) Y =-X +2       -2 = -(4) + 2    -2 = -2     Y = 4X -2      -2 = 4(4) - 2     -2 = 16 - 2    -2 = 14 A does not work. B) Y = -X + 2       -2 = -(4) + 2    -2 = -2      Y= -2 +4        -2 = -2 + 4       -2 = 2 B does not work C) Y = X -2         -2 = 4 - 2      -2 = 2       Y = X-6        -2 = 4 - 6       -2 = -2 C does not work D) Y =2X -10       -2 = 2(4) - 10    -2 = 8 - 10    -2 = -2       Y = X -6        -2 = 4 - 6          -2 = -2 D works, so the point (4, -2) is the solution to the system of equations in D
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