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Topic: **solve homework****Question:**
My little cousin needs help with algebra 1 homework and when i was in 8th grade i don't remember doing this. (Solving by eliminating) -x+Y=5
X-5Y=-9 ..I cant help him! SOMEONE PLEASE HELP ME!!!
i figured out Y that you!!! rita the dog, but i cant find x...
Never mind he figured it out :D THANK YOUI!!

July 18, 2019 / By Aiah

Add the two equations and the x's will drop out. Then you will have a single equation for y. Solve it. Once you have a number for y plug it into either equation to get a single equation for x. Solve it.

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To cancel the horrible fractions, you can multiply everything by 20, the lowest common multiple. This will give: 5x + 8 = 30 Subtract 8 from both sides: 5x +8 -8 = 30-8 = 5x = 22 divide both sides by 5 x = 22/5 To test it, just substitute 22/5 into the original equation 1/4(22/5) +2/5 = 1.1 + 0.4 = 1.5 = 3/2

first, you should find a common denominator for the fractions. to do this, you need to find the least common multiple that all 3 numbers go into which would be 20. (1/4)x+(2/5)=(3/2) (5/20)x+(8/20)=(30/20) ..........-(8/20)..-(8/20) (5/20)x=(22/20) *20........*20 5x=22 x=22/5 to make sure you're correct, plug the solution you got back into the original equation (1/4)(22/5) + (2/5) = (3/2) plug in your solution (11/10)+(4/10)=(15/10) find the common denominator: 10 and solve (15/10)=(15/10) correct :)

First step is to get x by itself. So we have to move 2/5 to the other side to get 1/4x = 3/2 - 2/5 The next step is to find a common denominator for all the fractions. 20 will work. 5/20x = 30/20 - 8/20 5/20x = 22/20 5x = 22 x = 22/5

–x + y = 5 Eq. (a million) x – 5y = –9 Eq. (2) –5x + 5y = 25 Eq. (a million) × 5 x – 5y = –9 Eq. (2) ------upload---------------------- –4x = sixteen x = sixteen/–4 = –4 –x + y = 5 Eq. (a million) x – 5y = –9 Eq. (2) ------upload----------------- –4y = –4 y = –4/–4 = a million (x, y) = (–4, a million)

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1.) 5x + 3y = 12 ; 4x - 5y = 17 the best way to answer this is by elimination method 5x + 3y = 12 --> multiply this by 5 4x - 5y = 17 --> multiply this by 3 25x + 15y = 60 12x -15y = 51 37x = 111 x= 3, then use substitution y = -1 2.) x + 2y = 6 ---> elimination method again (multiply this by -1) x - 3y = -4 -x -2y = -6 x -3y = -4 -5y = -10 y=2, if y =2 then x = 2 3. 5m + 2n = -8 elimination method(multiply this by -3) 4m + 3n = 2 ---->multiply this by 2 -15m -6n = 24 8m + 6n = 4 -7m = 28 m = -4, n = 6 ---------------------------------- x = 3 - 2y ; 2x + 4y = 6 2(3 - 2y) + 4y = 6 6 -4y + 4y = 6 6 = 6... this only means that this equation has infinite number of solutions

Unless you are told to use a specific method, I usually use elimination unless the value of one of the variables is very obvious. It really use your preference.

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