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Solve, using linear combination. 3x + y = 4 2x + y = 5

What shape is created if you connect the following points: (1,1), (5,1),(7,3), and (3,3) 2x + 2y = 5 3x + y = 5 solve using linear combination what is 6.4 - y = 6 2/5 solve the system using the linear combination method: My name is cristal and i need help on this linear combination 2x+y+-4 4x-2y=8 solve in linear combination 16x-5y=-33 & 16x+y=-51 solve 5x-4y+4z=18 -x+3y-2z=0 4x-2y+7z=3 use linear combination solving linear systems combination? using the substitiution method in linear equations how do you solve y=6x-5 y=-x+9 how to solve 3x-2y=12 with linear equation in t

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Solve, using linear combination. 3x + y = 4. 2x + y = 5. A ...


Solve, using linear combination. 3x + y = 4. 2x + y = 5. ... using linear combination. 3x + y = 4. 2x + y = 5. ... How to do the foil method to solve (2r+3s)(2r-s)
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3x+y=4 2x+y=5 solve using linear combination - Brainly.com


3x+y=4 2x+y=5 solve using linear combination 1. Ask for details ; Follow; Report; by vic99 03/01/2014. Log in to add a comment Verified answer Show verified answer ...
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SOLUTION: Solve, using linear combination. 3x + y = 4 2x + y = 5


... using linear combination. 3x + y = 4 2x + y = 5 Log On Ad: ... Linear Solvers Linear. ... Question 829473: Solve, using linear combination. 3x + y = 4
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3x+y=4 2x+y=5 solve using linear combinations


Answer to 3x+y=4 2x+y=5 solve using linear combinations ... Any help would be appreciated :) 1. List your company's revenue streams.
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Solve, using linear combination. 3x + y = 4 2x +... - OpenStudy


... using linear combination. 3x + y = 4 2x + y = 53x + y =4 2x +y = 5 then y=5-2x combine 2 expression 3x + ... using linear combination. 3x + y = 4 2x + y = 5.
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Solve, using linear combination. 3x + y = 4 2x + y = 5

line 1: 3x+y=4    or y=4-3x line 2: 2x+y=5 line 1 -line2...(3x-2x)+(y-y)=4-5 or x=-1 y=4-3x=4+3=7
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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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x-2y=.9 4x-3y=13.1

x-2y=.9 4x-3y=13.1 solving linear systems algebraically using substitution method Solve the first equation for x in terms of y.  x- 2y = 0.9 x = 2y + 0.9 Substitute the right-hand side of that equation for the x in the second equation and solve for y. 4x - 3y = 13.1 4(2y + 0.9) - 3y = 13.1 8y + 366 - 3y = 13.1 8y - 3y = 13.1 - 3.6 5y = 9.5 y = 9.5 / 5 y = 1.9 Substitute that for the y in the first equation and solve for x. x - 2y = 0.9 x - 2(1.9) = 0.9 x - 3.8 = 0.9 x = 4.7 Use both of those values in the second equation to check your work. 4x - 3y = 13.1 4(4.7) - 3(1.9) = 13.1 18.8 - 5.7 = 13.1 13.1 = 13.1
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solving linear systems using substitution

2x+y=-2 x-2y=4, so x=2y+4 2(2y+4)+y=-2 4y+y+8=-2 5y=-10 y=-5 x=(-y -2)/2  = (5-2)/2=3/2
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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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I need help, I really need help pleeease.

I assume that (3) is a graph, because it doesn't display on my tablet. Question 1 a) 1 because the daily operating cost is stated specifically. b) 3 because the profit of $500 will be a point on the graph corresponding to a value of t; 2 because if we put P=500 and solve for t we get t=1525/7.5=203.3 so the graph would show t=203-204. c) 2 because we set P=0 and solve for t=1025/7.5=136.7, so 137 tickets sold would give a profit of $2.5, but 136 tickets would make a loss of $5; alternatively, if (3) is a graph then P=0 is the t-axis so it's where the line cuts the axis between 136 and 137. d) The rate of change is 7.5 from (2).  e) 2, because the format shows it to be a linear relationship; a straight line graph for (3) also shows linearity. Question 2 a) profit=sales- operating costs so 1025 in (2) is the negative value representing these costs. If (3) is a graph, it's the intercept on the P axis at P=-1025; (4) is the P value when t=0. b) For (1) you would need to find out how many tickets at $7.50 you would need to cover the operating costs of $1025 plus the profit of $500. That is, how many tickets make $1525? Divide 1525 by 7.5; 2 and 3 have already been dealt with; to use (4) you would note that P=$500 somewhere between t=200 and 250 in the table. c) For (1), work out how many tickets cover the operating costs. 1025/7.5=136.7, so pick 137 which gives the smallest profit to break even; 2 and 3 already given; in (4) it's where P goes from negative to positive, between 100 and 150. d) For (1) the only changing factor is the number of tickets sold. The rate is simply the price of the ticket, $7.50; if (3) is a graph, the rate of change is the slope of the graph, to find it make a right-angled triangle using part of the line as the hypotenuse, then the ratio of the vertical side (P range) and the horizontal side (t range) is the slope=rate of change; for (4) take two P values and subtract the smallest from the biggest, then take the corresponding t values and subtract them, and finally divide the two differences to give the rate of change: example: (475-(-275))/(200-100)=750/100=7.5. e) For (1) it's clear that the profit increases (or loss decreases) with the sale of each ticket by the same amount as the price of a ticket, so there is a linear relationship; 2 and 3 already dealt with; in the table in (4) the profit changes by the fixed value of 50 tickets=$375, showing that a linear relationship applies between P and t: for every 50 tickets we just add $375 to the profit.
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solve for x and y in x+y=8

Problem: solve for x and y in x+y=8 this will be used to solve measurements in a geometry question. What you have there amounts to a linear equation: x + y = 8 x + y - x = -x + 8 y = -x + 8 Graphing this produces a straight (linear) line. You can choose any value for x, plug it into the equation, and get the corresponding value for y. If you were to choose x = 0, y would be 8. That is how we find the y-intercept on the graph. (0, 8) If you choose 8 for x, y would be 0. y = -8 + 8 (The negative represents -1, the slope of the line.)
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solve for x when (.182+x)/(.106-2x)=3.24

solve for x when (.182+x)/(.106-2x)=3.24 Im trying to solve for x its for a chemistry equation but i need to solve for x using the quadratic equation and i dont know which equals a b and c to find x. ************************** You can't solve this with the quadratic formula, because there is no x^2 term. Observe: (.182+x)/(.106-2x)=3.24 Multiply both sides by (.106-2x) to eliminate the fraction on the left. (.106-2x) * (.182+x)/(.106-2x) = (.106-2x) * 3.24 0.182 + x = 0.34344 - 6.48x This is a linear equation. Even if you arranged it so that all the terms are on the left, leaving a zero on the right side, you still would not have an x^2 term. That means that the "a" coefficient would be zero. When you try to use that in the quadratic formula, the denominator would be 2a = 2 * 0 = 0. You cannot divide by zero, so you are now stuck, with no possible solution. If you solve the equation from where I have taken it, you will find the value of x. 0.182 + x = 0.34344 - 6.48x 6.48x + x = 0.34344 - 0.182 7.48x = 0.16144 (Note that you could now subtract 0.16144 from both sides to get 7.48x - 0.16144 = 0, but you still don't have an x^2 term, so a=0, b=7.48 and c=-0.16144. That pesky 0 for the value of a will give that 0 denominator. Impossible to divide.) x = 0.0215828877
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substitution method for 6x-3y=18, and y=2x-6, and show steps

substitution method for 6x-3y=18, and y=2x-6, and show steps I need assistance in answering the question; how would I use the substitution method to solve the system linear equation of 6x-3y=18, and y=2x-6 In the second equation, you have the value of y in terms of x. Substutute that into the first equation in place of the y. 6x - 3y = 18 6x - 3(2x - 6) = 18 Now, you have an equation with only one unknown, x. Solve the equation. 6x - 3(2x - 6) = 18 6x - 6x + 18 = 18 18 = 18 Theoretically, this implies that you can have any value for x, but zero is the best choice here. Substitute zero for x into equation 1. 6x - 3y = 18 6(0) - 3y = 18 0 - 3y = 18 -3y = 18 y = 18/-3 y = -6 Re-work equation 1 so it is in slope/intercept form. 6x - 3y = 18 -3y = -6x + 18 y = 2x - 6 From this, you can see that the y-intercept is -6. This occurs when x is zero.
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solve a system of linear equations with three variables:

solve a system of linear equations with three variables: x+y-3z=1 2x-y+z=9 3x+y-4z=8 1) x + y - 3z = 1 2) 2x - y + z = 9 3) 3x + y - 4z = 8 Multiply equation 2 by 3. 3(2x - y + z) = 9 * 3 4) 6x - 3y + 3z = 27 Add equation 4 to equation 1.      x +   y - 3z =   1 +(6x - 3y + 3z = 27) --------------------------    7x - 2y        = 28 5) 7x - 2y = 28 Multiply equation 2 by 4. 4(2x - y + z) = 9 * 4 6) 8x - 4y + 4z = 36 Add equation 3 to equation 6.    8x - 4y + 4z = 36 +(3x +  y - 4z =   8) ------------------------   11x - 3y       = 44 7) 11x - 3y = 44 Multiply equation 5 by 3. 3(7x - 2y) = 28 * 3 8) 21x - 6y = 84 Multiply equation 7 by 2. 2(11x - 3y) = 44 * 2 9) 22x - 6y = 88 Subtract equation 9 from equation 8.    21x - 6y = 84 -(22x - 6y = 88) ---------------------      -x        =  -4 x = 4  <<<<<<<<<<<<<<<<< Use equation 5 to solve for y. 7x - 2y = 28 7(4) - 2y = 28 28 - 2y = 28 -2y = 0 y = 0  <<<<<<<<<<<<<<<<< Use equation 3 to solve for z. 3x + y - 4z = 8 3(4) + 0 - 4z = 8 12 - 4z = 8 -4z = -4 z = 1  <<<<<<<<<<<<<<<<< Check your answers. 1) x + y - 3z = 1    4 + 0 - 3(1) = 1    4 - 3 = 1    1 = 1 2) 2x - y + z = 9    2(4) - 0 + 1 = 9    8 + 1 = 9    9 = 9 3) 3x + y - 4z = 8    3(4) + 0 - 4(1) = 8    12 - 4 = 8    8 = 8 Everything checks. Answer: x = 4, y = 0, z = 1  
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