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x + 3y = 7 x - 3y = 1 Solve the system of equations.

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Solve the system of equations x + 3y = 7 x - 3y = 1?


Solve the system of equations x + 3y = 7 x - 3y = 1? Find answers now! ... 1. Solve the system: 2x + 3y = 7 6x ... the system: x + y = 9 y + z = 7 x - z = 2 . 4 ...
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x + 3y = 7, 3x - 6y = 1 - System of Equations Calculator ...


Free system of equations calculator - solve system of equations step-by-step. Symbolab; Solutions Graphing Calculator ... system-of-equations-calculator. x + 3y = 7 ...
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x + 3y = 7 x - 3y = 1 Solve the system of equations. A) x = 4 ...


x + 3y = 7 x - 3y = 1 Solve the system of equations. A) x = 4, y = 1 B ... 3y - 3y = 1 7 -6y = 1-7 -7-6y = -6 y = 1 Now solve for x X + 3y = 7 x + 3 = 7 x = 4 Now ...
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x + 3y = 7 x - 3y = 1 Solve the system of equ... - OpenStudy


x + 3y = 7 x - 3y = 1 Solve the system of equations. A) ... Solve x+3y=7;x−3y=1 Steps: ... I will try to solve your system of equations. x+3y=7;x−3y=1 Step: ...
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solve the following system of linear equations using ... - eNotes


Solve the system of equations by substitution: ... (1) x-3y=7 (2) Solve (2) for x. x=7+3y (3) Substitute ... eNotes.com is a resource used daily by thousands of ...
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How do solve the following linear system?: x - 3y = 8 , 3y ...


The solution for the system of equations is : color(blue)(x=23/7, y ... How do solve the following linear system?: # x - 3y = 8 ... How do you solve #x=y+4# and #x=2y ...
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Solve system of equations x+3y=7 (1) x=8-3y (2) no solution


Solve system of equations x+3y=7 (1) x=8-3y (2) no solution. 56,197 results. ALGEBRA ... Solve the system x+4y=1,-x+y=4.What is the -coordinate of this solution x.
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How do you solve the system using the elimination ... - Socratic


How do you solve the system using the elimination method ... In elimination method in solving simultaneous Equations; ... How do you solve the system #x+y=4#, #2x+3y=0#?
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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 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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solve the system by substitution

solve the system by substitution solve the system by substitution x=4y-11 -3x+4z=-7 y=-5x+2z+25 almost made it but got stuck at the end. 1) x = 4y - 11 2) -3x + 4z = -7 3) y = -5x + 2z + 25 Equation 3 already has y in terms of x, so substitute that for y in equation 1. x = 4y - 11 x = 4(-5x + 2z + 25) - 11 x = -20x + 8z + 100 - 11 x + 20x - 8z = -20x + 8z + 89 + 20x - 8z 4) 21x - 8z = 89 Now, there are two equations with x and z. Solve equation 2 for x and substitute that into equation 4. -3x + 4z = -7 -3x + 4z - 4z = -7 - 4z -3x = -7 - 4z x / -3 = -7/-3 - (4/-3)z x = 7/3 + 4/3z 21x - 8z = 89 21(7/3 + 4/3z) - 8z = 89 49 + 28z -8z = 89 49 + 20z - 49 = 89 - 49 20z = 40 20z / 20 = 40 / 20 z = 2   <<<<<<<<<<<<<<<<<< Plug that into equation 2 and solve for x. -3x + 4z = -7 -3x + 4(2) = -7 -3x + 8 = -7 -3x + 8 - 8 = -7 - 8 -3x = -15 -3x / -3 = -15 / -3 x = 5   <<<<<<<<<<<<<<<<<< Plug the value of x into equation 1 and solve for y. x = 4y - 11 5 = 4y - 11 5 + 11 = 4y - 11 + 11 16 = 4y 4y = 16 4y / 4 = 16 / 4 y = 4   <<<<<<<<<<<<<<<<<< Use all three of the original equations to check your answers. 1) x = 4y - 11    5 = 4(4) - 11    5 = 16 - 11    5 = 5 2) -3x + 4z = -7    -3(5) + 4(2) = -7    -15 + 8 = -7    -7 = -7 3) y = -5x + 2z + 25    4 = -5(5) + 2(2) + 25    4 = -25 + 4 + 25    4 = 4 Answer: x = 5, y = 4, z = 2  
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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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word problem please show how

Suppose that in one week, your company takes in $2,220 in sales at one of its cell phone stores. The price breakdown per phone is as follows: Cell phone model A4: $35 Smartphone Z20: $50 Suppose that a total of 51 phones were sold. Set up the system of equations that needs to be solved to determine how many of each type of phone were sold. Give a clear definition of the variables in the system. Solve the system of equations, showing clearly how the solution was determined, and state the results clearly in light of the real-world situation. Verify your results of the 2 linear equations by graphing in the desired graphing program and paste the graph in your assignment document (edit/copy image). Explain how the results are verified by the graph.
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i need the answer for these questions

Part 1 Newton’s Method for Vector-Valued Functions Our system of equations is, f1(x,y,z) = 0 f2(x,y,z) = 0 f3(x,y,z) = 0 with, f1(x,y,z) = xyz – x^2 + y^2 – 1.34 f2(x,y,z) = xy –z^2 – 0.09 f3(x,y,z) = e^x + e^y + z – 0.41 we can think of (x,y,z) as a vector x and (f1,f2,f3) as a vector-valued function f. With this notation, we can write the system of equations as, f(x) = 0 i.e. we wish to find a vector x that makes the vector function f equal to the zero vector. Linear Approximation for Vector Functions In the single variable case, Newton’s method was derived by considering the linear approximation of the function f at the initial guess x0. From Calculus, the following is the linear approximation of f at x0, for vectors and vector-valued functions: f(x) ≈ f(x0) + Df(x0)(x − x0). Here Df(x0) is a 3 × 3 matrix whose entries are the various partial derivative of the components of f. Specifically,     ∂f1/ ∂x (x0) ∂f1/ ∂y (x0) ∂f1/ ∂z (x0) Df(x0) = ∂f2/ ∂x (x0) ∂f2/ ∂y (x0) ∂f2/ ∂z (x0)     ∂f3/ ∂x (x0) ∂f3/ ∂y (x0) ∂f3/ ∂z (x0)   Newton’s Method We wish to find x that makes f equal to the zero vector, so let’s choose x = x1 so that f(x0) + Df(x0)(x1 − x0) = f(x) =  0. Since Df(x0)) is a square matrix, we can solve this equation by x1 = x0 − (Df(x0))^(−1)f(x0), provided that the inverse exists. The formula is the vector equivalent of the Newton’s method formula for single variable functions. However, in practice we never use the inverse of a matrix for computations, so we cannot use this formula directly. Rather, we can do the following. First solve the equation Df(x0)∆x = −f(x0) Since Df(x0) is a known matrix and −f(x0) is a known vector, this equation is just a system of linear equations, which can be solved efficiently and accurately. Once we have the solution vector ∆x, we can obtain our improved estimate x1 by x1 = x0 + ∆x. For subsequent steps, we have the following process: • Solve Df(xi)∆x = −f(xi). • Let xi+1 = xi + ∆x
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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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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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solve the system of equations by graphing 9x-4y=12 and 9x=4y+12

Problem: solve the system of equations by graphing 9x-4y=12 and 9x=4y+12 solve the system of equations by graphing 9x-4y=12 and 9x=4y+12 Both equations are for the same line!
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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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