how do you solve x over 5 equals x over 4 minus 1 as a linear equation?


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how do you solve x over 5 equals x over 4 minus 1 as a linear equation?

i need to know this so that i can know how to solve problems without help next time

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How do you solve the equation x over 5 equals 4 - Answers.com


How do you solve the equation x over 5 equals 4? ... Can anyone solve x over 4 minus 4 equals 3x over 4 ... How do you solve the equation for x -4 over 3x equals 1 ...
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Solving One-Step Linear Equations (page 2 of 4) - Purplemath


Solving One-Step Linear ... I multiplied by 5 on the right-hand side of the equation, and by 5/1 on ... by putting it over 1. Then the solution is x ...
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Solve for x x over 4 - 5 over 8 equals x plus 316 - Answers.com


Solve for x x over 4 - 5 over 8 equals x ... How do you solve the equation for x -4 over 3x ... The quantity of x plus 1 over 2 plus the quantity of 4x minus 1 over 4 ...
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5 Ways to Solve for X - wikiHow


There are a number of ways to solve for x, whether you're ... wiki How to Solve for X. Five Methods: Using a Basic Linear Equation ... You'll get x left over on ...
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Solving Linear Equations: 'One-Step' Equations


Solving One-Step Linear Equations (page 1 of 4) Sections ... x plus nothing is x, 6 minus ... You can use the Mathway widget below to practice solving a linear ...
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How do you solve five equals x2 plus six x? - Weknowtheanswer


How do you solve five equals x2 plus six x? ... how do you solve x over 5 equals x over 4 minus 1 as a linear equation? ... twelve x minus five over six x plus 3 ...
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how do i solve 3 over x equals 6 over 12 - search results


how do i solve 3 over x equals 6 over 12 ... How do I solve this equation? 1+1/x over x over 1/x-1 By the way this ... Solve each equation: 5 over 2x-5 minus 4/5= 1 ...
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Solve for x: 2 over 5 plus 3 over 5 x equals quantity x plus ...


Solve for x: 2 over 5 plus 3 over 5 x equals quantity x plus 5 ... (x+5) distribute 4x+6=x^2+5x minus ... How can the properties of linear pairs and vertical angles ...
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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
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how do you solve for x in this equation with these fractions x/2+4=5/2x-6

Problem: how do you solve for x in this equation with these fractions x/2+4=5/2x-6 the equation includes fractions and variables on both sides. the equation is x over 2 plus 4 equals 5 over 2 with x adjacent to the fraaction minus 6. solving for x. Problems are easier to deal with if we can elim
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how to solve -4/9(2x-4)=48

Is this -4/(9(2x-4))=48? Or (-4/9)(2x-4)=48? We can find the answers to both questions. Let's take the first interpretation first and see what we get. To solve it we need to get rid of the fraction. In all questions of this type we take the denominator of the fraction (the bottom bit), and mul
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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
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how do you solve x over 5 equals x over 4 minus 1 as a linear equation?

(x/5)=(x/4)-1...mult both sides bi 20 4x=5x-20 -x=-20 x=20
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How do i solve simultaneous equation 5x+9y=-8 and x-y=10 in a graph?

Ok.  The graphing method is really easy.  Your equations are easy.  I'll tell you as cleary as possible how to graph it.  Ok.  First, you have to solve for each variable for each equation.  You can solve for y if x = 0, and solve for x if y = 0.  The answers w
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please show step by step to solve the following equation 5x + 7-3x =17-9x+12

Collect numbers on one side of the equation and variables (and multiples of variables) on the other side. That means moving things across the equals sign. Equals changes plus to minus and minus to plus when a transfer takes place from side to the other. So if we transfer all x's to left we get
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how can I simplify the last problem

26r-2=3r^2. We need to put this into standard form. The standard quadratic form is ax^2+bx+c=0, where x is the unknown and a, b and c are numbers. The unknown is r in this problem, rather than x, so let's put the equation into quadratic form: 3r^2-26r+2=0. How did I get this? Subtract 26r
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3x squared minus 7x equals 0

3x squared minus 7x equals 0 in my math homework...solve the quadratic equations by factoring 3x^2 - 7x = 0 When we factor this, we will have: (3x + a)(x + b) = 0 We need an a and b such that 3b + a = -7, and ab = 0. We know ab has to equal zero, because there is no constant term follow
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A temperature of 59F equals 15C. A temperature of 68F equals 20C.

Well!!!   Let us assume  A F + B = C ---- {1}   59F equals 15C ==> 59 A + B = 15 ......{2} 68F equals 20C ==> 68 A + B = 20.... {3}   Solving {2} and {3}, {3} - {2} ==> 9 A = 5 ==> A = 5/9 {2} ==> B = 15 - 59 A = 15 - 59 (5/9) =
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