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Topic: Algebra 1 Question Please Help! 10pts?**Question:**
1.) 2x-y=3 & 5x-2y=10
i need to solve that problem by using the ADDITION METHOD. Can someone please help?

June 26, 2019 / By Aston

i know you are asking about the value of x & y;; right? this is how to solve that; step 1: 2x-y=3 (eq.1) 5x-2y=10 (eq.2) step 2: using (eq.1); we have 2x-y=3 ; find the y; y=2x-3 (eq.3) step 3: using (eq.2); we have 5x-2y=10 (eq.2) ; find y; 5x-10=2y y=(5x-10)/2 (eq.4) step 4: equating (eq.3) & (eq.4) 2x-3 = (5x-10)/2 4x-6=5x-10 4x-5x=-10+6 -x = -4 x = 4................answer step 5: using (eq.3) we have; y=2x-3 ; substituting x=4; we have y=2(4)-3 y= 5.............answer thats it EngrTOP20

👍 112 | 👎 2

Did you like the answer? First of all: what was the question asking you to do with equation? well im just going to assume that they wanted you to simplify so this is how you do it: first you combine all the terms that are being multiplied by x and that equals -12x then you combine all the terms that are just numbers, which is equal to 1 after that, you comkbine all the x 2 terms and that put together is equal to -10x 2 and with the x 3 term, there is no other term that is with the exponent to the third so you leave that alone, so it is -x 3 after that is all finished, you put the whole equation together to equal -12x+1-10x 2-x 3 and if you are still having trouble, feel free to consult a math teacher. most of the time they are more than happy to help you

Call your 2 equations (1) and (2) 2x - y = 3 (1) 5x - 2y =10 (2) multiply equation (1) by -2 to get -4x + 2y = -6 Call this equation (3) Now add equations (2) and (3) to get x = 4 now plug x into any of your equations and solve for y e.g. in equation (1) 2 (4) - y = 3 8-y=3 -y=-5 so y = 5 final answer: x=4, y=5

👍 40 | 👎 -5

2x - y = 3, multiply by - 2 the equation 5x - 2y = 10 - 4x + 2y = - 6 5x - 2y = 10 ---------------------- x = 4 2x - y = 3, multiply by 5 5x - 2y = 10, multiply by - 2 10x - 5y = 15 -10x + 4y = - 20 ------------------------ - y = - 5 y = 5

👍 39 | 👎 -12

Width = w Length = three+w Perimeter of a rectangle = 58cm = two(L+W) ==> two[(three+w)+w] = fifty eight ==> two[three+2w] = fifty eight ==> three+2w = fifty eight/two = 29 ==> 2w = 29 - three = 26 ==> w = 26/two = 13cm Therefore, duration = three+thirteen = 16cm

👍 38 | 👎 -19

- 10x + 5y = - 15 10x - 4y = 20---------ADD y = 5 2x - 5 = 3 2x = 8 x = 4 x = 4 , y = 5

👍 37 | 👎 -26

Whenever there are fractions as coefficients, I teach my algebra classes to multiply through by a common denominator, then solve as normal. So, in question 1, a common denominator is 15. So mutliply EACH term by 15. This results in 15(1/3x) + 15(2/5x) = 15(2) (remember you must do same thing to all terms to maintain equality. Simplifying, you get 5x + 6x = 30 Combining like terms: 11x = 30 dividing by 11 on both sides, you get x = 30/11 Same concept for problem 2, with a common denominator of 12 so: 12 (1/3(x+2)) = 12 (1/4(2x-4)) ==> 4(x+2) = 3(2x - 4) (simplifying) ==> 4x + 8 = 6x - 12 (distributive property) ==> 20 = 2x (getting like terms on same side and simplifying) ==> 10 = x (dividing both sides by 2) Hope this helps

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