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how do you solve 3/x+5

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5 Ways to Solve for X - wikiHow


How to Solve for X. There are a number of ways to solve for x, whether you're working with exponents and radicals or if you just have to do some division or ...
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How do you solve 3(x+ 5) +x=7? | Socratic


How do you solve #3(x+ 5) +x=7#? Algebra. 2 Answers ? 1 Tony B ...
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How do you solve -3(x-5)<9? | Socratic


How do you solve #x+12<=5# or #3x-21>=0#? How do you solve and graph #5-5x>4(3-x)#? How do you solve and graph #z+3>2/3#? See more ...
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How to solve [math]3^{2-x} = 5^{x+4}[/math] - Quora


How do you solve [math] 3^x + 4^x = 5^x [/math]? How do I solve [math]x^6-x^5+3x^4-x^2-3=0[/math]? How do I solve for x in |x-3|+2|x+1|=4? How many solutions are ...
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How do you solve [math] 3^x + 4^x = 5^x [/math] - Quora


How do you solve [math] 3^x + 4^x = 5^x [/math]? Update Cancel. Promoted by The Great Courses Plus. Want to take your math skills to the next level?
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Solving Exponential Equations from the Definition | Purplemath


Demonstrates how to solve exponential equations by using the definition of exponentials, ... Solve 5 x = 5 3; Since the bases ("5" in each case) are the same, ...
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How do you Solve: 3x^2-2x=-5? | eNotes


How to solve this expression? `(5/6)^(-3)[(5/6)^(3)-:(36/25)^(-2)]^(3){-[(-6/5)^(2)]^(-4)}^(3)=... How do you solve 5-1/2(x-6)=4; More Math Questions. Popular Questions.
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Solving One-Step Linear Equations - Purplemath


Solving One-Step Linear Equations (page 2 of 4) Sections ... Solve 3 / 5 x = 10; Since x is multiplied by 3 / 5, I'll want to multiply both sides by 5 / 3, ...
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Solving Rational Equations, Ex 3 - Khan Academy


Practice: Equations with one rational expression (advanced) ... or any of the ways that we know how to solve quadratics. So let's do that.
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How do you solve this problem: x/3-8=2. What do ... - Brainly.com


How do you solve this problem: x/3-8=2. What do you get for x step by step please. 1. Ask for details ; Follow; Report; by angie8860 01/17/2017. Log in to add a comment
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x^2+2x-15=0

x^2 + 2x - 15 = 0 To solve this equation (i.e. to find the value of x), you need to factorise it: To factorise an equation you need to look at the multiples of the number without any 'x'. In your equation this is -15. So multiples of 15 are: 1x15, 3x5. You then choose the multiples that will equal the term with the 'x' , when added. To do this you need to take into account the fact that for your equation the last term is negative. So the only possible choice for this equation is 3 and 5 (as the term '2x' can never be gained by adding or subtracting 1 and 15). And so now (because we need -15) the only possible choices are: (x - 3)(x + 5)      or     (x + 3)(x - 5) Now if you imagine multiplying out the brackets you can see that the only choice that is going to give the right answer is:     (x - 3)(x + 5)    And so x^2 + 2x - 15 = (x - 3)(x + 5) = 0 And so now you can see that the only times when this equation is going to equal zero is when:     x = 3    or   x = -5     , and this is the final solution to the equation.
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solve the equation ln(3x 5)-1=ln(2x-3)

???????????? wot cha meen bi "(3x5)" iz zat 3*x^5 ????????? or did yu  miss hit  kee & it shood be sumthun like "(3x+5)" ????????????????
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if I want to make a 3x5 index card 15 inches longer, how wide would it be?

?? maebee yu tri tu impli yu wanna make kard BIGGER & it keep the same shape?????? 1 side from 5 tu 20, so nu leng=4*old-leng maebee yu want nu-wide=4*old-wide??? 4*3=12
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Solve 6(3 a+4)+5(4 a -2); a=5

Let f(a)=6(3a+4)+5(4a-2)···Eq.1   Expand brackets: f(a)=18a+24+20a-10   Combine like terms: f(a)=38a+14   Substitute a=5 for a in the equation: f(5)=38x5+14=190+14=204 CK: Plug a=5 into Eq.1: 6(3x5+4)+5(4x5-2)=6x19+5x18=114+90=204  CKD. The answer is 204
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Shawn took in $69.15 in two hours work on Saturday selling Belts for $8.05 and earrings for $4.50. How many of each did Shawn sell?

Shawn took in $69.15 in two hours work on Saturday selling Belts for $8.05 and earrings for $4.50. How many of each did Shawn sell? I'm not too sure how ​you are expected to solve this problem. You will end up with a Diophantine Equation (one involving integer coefficients and integer solutions only) If you have never heard of Diophantus, then the precise mathematical solution later on should be ignored. Instead ...   The (Diophatine) equation is 161B + 90E = 1363 You should be able to show that Shawn can only sell a maximum of 8 belts (i.e. if he sells no earrings), and a maximum of 15 earrings (i.e. if he sells no belts). So start with the smaller number (8). Shawn sells up to 8 belts. So put B = 1 in the (Diophantine ) equation and see if B is an integer. If not, then B=1 cannot be a solution, So now try again with B = 2, and so on ..     My precise mathematical solution now follows. Let B be the number of belts sold @ $8.05 per belt. Let E be the number of earrings sold @ $4.50 per pair. Total money made is P = $69.15 Total money made is: B*$8.05 + E*$4.50 Equating the two expressions, 8.05B + 4.50E = 69.15 Making the coefficients as integers we end up with a Diophantine equation. 161B + 90E = 1383 We now manipulate the coefficients 161 and 90. We should end up with a relation between the two that should produce the value 1. This manipulation is in two parts. In the first line, we are writing the larger coefft, 161, as a multiple of the smaller coefft plus a remainder. 1st Part 161 = 1x90 + 71  90 = 1x71 + 19     now we write the 1st multiple as a multiple of the 1st remainder + remainder 71 = 3x19 + 14    we write the 2nd multiple as a multiple of the 2nd remainder + remainder 19 = 1x14 + 5 14 = 2x5 + 4 5 = 1x4 + 1 ============ 2nd Part We rewrite that last equation so that 1 is on the left hand side 1 = 5 – 1x4 We now use the remainder from the next equation up. 1 = 5 – 1x(14 – 2x5) 1 = 3x5 – 1x14 And again we use the remainder in the next equation up 1 = 3(19 – 1x14) – 1x14 1 = 3x19 – 4x14 1 = 3x19 – 4(71 – 3x19) 1 = 15x19 – 4x71 1 = 15(90 – 1x71) – 4x71 1 = 15x90 – 19x71 1 = 15x90 – 19(161 – 1x90) 1 = 34x90 – 19x161 ================ Therefore, 1383 = (-19x1383).161 + (34x1383).90 1383 = -26,277*161 + 57,022*90 Which implies B = -26,277 E = 47,022 which is obviously incorrect! But all is not lost. These two values for B and E are simply two values that happen to satisfy the original Diophantine equation. What we need is the general solution, which now follows. In general, B = -26,277 + 90k    (using the coefft of E) E = 47,022 – 161k    (using the coefft of B) If you substitute these expressions into the original Diophantine equation, the k-terms will cancel out, leaving you with the original eqn. We know that B and E must be smallish numbers and that neither can be negative. For example, if no earrings were sold, Shawn would need to sell over 8 belts to clear $69, and if no belts were sold, then Shawn would need to sell over 15 earrings. So Shawn needs to sell between 0 and 8 belts and between 0 and 15 earrings. Setting k = 292, B = -26,277 + 90*292 = 3 E = 47,022 – 161*292 = 10 B = 3, E = 10 If k is greater than or less than 292 by 1, or more, then either B will be negative or E will be negative. So this is the answer. Check 3*8.05 + 10*4.50 = 25.15 + 45.00 = 69.15 – Correct! Answer: Shawn sells 3 belts and 10 earrings
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How do I solve y=-x-4

How do I solve y=-x-4 I need to be able to plot these points on a graph but i don't know how to solve the problem.   make an x & y axis. ( like a cross one line goes vertical an the other horizontal) put equal dimention for x & y line From center line moving to the left is negative line moving down is negative Just assign values for x & y then locate it at the x & y axis.                                                                                                                                                            SOLVE Y                                                                                        POINTS X Y 1 0 SOLVE Y 2 SOLVE X 0 3 1 SOLVE Y 4 SOLVE X 1 5 2 SOLVE Y 6 SOLVE X 2 7 -1 8 SOLVE X -1 9 -2 SOLVE Y 10 SOLVE X -2 11 3 SOLVE Y y=-x-4 sample: in point 1 => if  x = 0 y = - (0) - 4 y = - 4 in point 2 => if y = 0 0 = - x - 4 x = - 4 in point 7 => if x=-1 y = -x -4 y =-(-1) -4 y = +1 - 4 y = - 3 at point 8 => if y = -1 -1 = -x -4 x = -4 + 1 x = - 3 - 3 POINTS X Y 1 0 - 4 2 - 4 0 3 1 - 5 4 - 5 1 5 2 - 6 6 - 6 2 7 -1 8 - 3 -1 9 -2 - 2 10 - 2 -2 11 3 - 7 if you have done the 1 to 4 then you can plot the points from 1 to 11 and there is your graph.  
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how to solve for x with fractions

Simplifying x3 + 3x2 + -4x = 0 Reorder the terms: -4x + 3x2 + x3 = 0 Solving -4x + 3x2 + x3 = 0 Solving for variable 'x'. Factor out the Greatest Common Factor (GCF), 'x'. x(-4 + 3x + x2) = 0 Factor a trinomial. x((-4 + -1x)(1 + -1x)) = 0 Subproblem 1 Set the factor 'x' equal to zero and attempt to solve: Simplifying x = 0 Solving x = 0 Move all terms containing x to the left, all other terms to the right. Simplifying x = 0 Subproblem 2 Set the factor '(-4 + -1x)' equal to zero and attempt to solve: Simplifying -4 + -1x = 0 Solving -4 + -1x = 0 Move all terms containing x to the left, all other terms to the right. Add '4' to each side of the equation. -4 + 4 + -1x = 0 + 4 Combine like terms: -4 + 4 = 0 0 + -1x = 0 + 4 -1x = 0 + 4 Combine like terms: 0 + 4 = 4 -1x = 4 Divide each side by '-1'. x = -4 Simplifying x = -4 Subproblem 3 Set the factor '(1 + -1x)' equal to zero and attempt to solve: Simplifying 1 + -1x = 0 Solving 1 + -1x = 0 Move all terms containing x to the left, all other terms to the right. Add '-1' to each side of the equation. 1 + -1 + -1x = 0 + -1 Combine like terms: 1 + -1 = 0 0 + -1x = 0 + -1 -1x = 0 + -1 Combine like terms: 0 + -1 = -1 -1x = -1 Divide each side by '-1'. x = 1 Simplifying x = 1 Solution x = {0, -4, 1}
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How do you solve this problem? solve for d: n= m+3d/4

Problem: How do you solve this problem? solve for d: n= m+3d/4 I need help solving this and seeing how it was solved since I am having trouble figuring out what to do and how to solve it thanks. n = m + 3d/4 n - m = (m + 3d/4) - m n - m = 3d/4 4(n - m) = (3d/4)4 4(n - m) = 3d 4(n - m)/3 = 3d/3 4/3 (n - m) = d Answer: d = 4/3 (n - m)  
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how can i solve these equations?

how can i solve these equations? x + 3y – z = 2 x – 2y + 3z = 7 x + 2y – 5z = –21 We eliminate one of the unknowns (x, y or z, take your choice), leaving two unknowns. Then, we eliminate a second one, giving us an equation with only one of the unknowns. Solve for that and plug that value into one of the equations to solve for a second unknown. Finally plug both of those values into an equation to solve for the third unknown. It sounds complicated, but if you follow a logical sequence, the problem solves itself. 1) x + 3y – z = 2 2) x – 2y + 3z = 7 3) x + 2y – 5z = –21 If we subtract equation 3 from equation 2, we eliminate the x.    x – 2y + 3z =     7 -(x + 2y – 5z = –21) ------------------------     - 4y  + 8z =   28 4) -4y + 8z = 28 Subtract equation 1 from equation 2, eliminating the x again.    x – 2y + 3z = 7 -(x + 3y –   z = 2) ----------------------      - 5y +  4z = 5 5) -5y + 4z = 5 You now have two equations with only a y and a z. The easiest step now is to eliminate the z. Multiply equation 5 by 2. 2 * (-5y + 4z) = 5 * 2 6) -10y + 8z = 10 Subtract equation 6 from equation 4, eliminating the z.     -4y + 8z = 28 -(-10y + 8z = 10) ---------------------      6y      =   18 6y = 18 y = 3  <<<<<<<<<<<<<<<<<<<<< Plug that into equation 5 to solve for z. -5y + 4z = 5 -5(3) + 4z = 5 -15 + 4z = 5 4z = 20 z = 5  <<<<<<<<<<<<<<<<<<<<< Plug the values of y and z into equation 1 to solve for x. x + 3y – z = 2 x + 3(3) – 5 = 2 x + 9 - 5 = 2 x + 4 = 2 x = -2  <<<<<<<<<<<<<<<<<<<<< Always check the answers by plugging all three values into one of the original equations. Using all three would be even better. Equation 2: x – 2y + 3z = 7 (-2) – 2(3) + 3(5) = 7 -2 - 6 + 15 = 7 -8 + 15 = 7 7 = 7 Equation 3: x + 2y – 5z = –21 (-2) + 2(3) – 5(5) = –21 -2 + 6 - 25 = -21 6 - 27 = -21 -21 = -21 Answer: x = -2, y = 3, z = 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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