Solve an elimination question please?

👨🏻‍🏫 學術台 數學
2008-06-16 8:01 am
4x+3y=7
2x=3y-5=0

Please go through step by step how to solve this. It's been stumping me for a good half hour.
更新1:

Oops, I meant to say 2x+3y-5=0! Thanks for pointing that out, I missed it ;)

回答 (10)

2008-06-16 8:17 am
✔ 最佳答案
4x+3y=7 ....name this eqn.......[a]
2x+3y-5=0 ==>2x+3y=5..........[b]
Now, [a] - [b] gives,
4x+3y - (2x+3y) = 7 -5
4x+3y -2x-3y = 2 (y term getting cancelled)
2x = 2
x = 1
From [a], 4*1 +3y =7
4 +3y=7 ==> 3y=7-4
3y = 3 ==> y=1
ANS: x = 1; y = 1
👍 0 👎 0
2008-06-17 11:52 pm
4x + 3y = 7
2x + 3y = 5

4x + 3y = 7
-4x - 6y = - 10----ADD

-3y = - 3
y = 1

2x + 3 = 5
2x = 2
x = 1

x = 1 , y = 1
👍 0 👎 0
2008-06-16 3:52 pm
4x + 3y = 7
2x + 3y - 5 = 0

2x + 3y - 5 = 0
2(2x + 3y - 5) = 2(0)
4x + 6y - 10 = 0
4x + 6y = 10

...4x + 3y = 7
--4x + 6y = 10
-------------------------
-3y = -3

-3y = -3
y = -3/-3
y = 1

4x + 3y = 7
4x + 3(1) = 7
4x + 3 = 7
4x = 7 - 3
4x = 4
x = 4/4
x = 1

∴ x = 1 , y = 1
👍 0 👎 0
2008-06-16 3:37 pm
solution
4x + 3y = 7
2x + 3y - 5 = 0

1st:
simplify the 2nd eq.
2x + 3y - 5 = 0
2x + 3y = 5

2nd:
sub. the 2 eq.
4x + 3y = 7
2x + 3y = 5
¯¯¯¯¯¯¯¯¯
2x = 2 ans. after subtracting

3rd:
solve for x.
2x = 2
x = 2 / 2
x = 1

4th:
substitute the value of x in any eq. (given eq.)

(if substitute at the 1st eq.)
x = 1
; 4x + 3y = 7
4 (1) + 3y = 7
4 + 3y = 7
3y = 7 - 4
3y = 3
y = 3 / 3
y = 1

therefore.... the solution is (1,1)
👍 0 👎 0
2008-06-16 3:30 pm
4x+3y = 7
2x+3y-5 = 0 ⇒ 2x+3y = 5

Subtract eq. 2 from eq. 1.
2x = 2
x = 1

4(x) + 3y = 7
y = 1
👍 0 👎 0
2008-06-16 3:20 pm
x=1
y=1

2x + 3y - 5 = 0 add 5 to both sides and you get..
2x + 3y = 5 if you put the other equation directly above
Then just cross out all the like parts of the equation like subtraction math

4x + 3y = 7
- (2x + 3y = 5)
___________
2x + 0y = 2
2x = 2
x = 2/2 = 1 x=1

put x = 1 into the equation. 4x + 3y = 7 and solve for y.

4(1) + 3y = 7
4 + 3y = 7
4 + 3y - 4 = 7 - 4 which equals 3y = 3
3y/3 = 3/3 = 1 so y = 1

x=y=1


so x=1
參考: the school of hardknocks. It could be wrong.
👍 0 👎 0
2008-06-16 3:20 pm
4x+3y=7
2x+3y=5

Now substract the two equations
We get x=1
substitute this value in any of the equation
We get y=1
👍 0 👎 0
2008-06-16 3:18 pm
First, 5 needs to be added to both sides of the second.
1. 4x+3y=7
2x+3y=5

Second, subtract the second from the first.
2. 4x+3y=7
- (2x+3y=5)
Which gives:
2x=2
Therefore, by dividing both sides by 2, x=1.

Third, substitute x=1 in for the x in either equation. For simplicity I'll use the second one.
3. 2(1)+3y-5=0
Which simplifies to:
-3+3y=0
Adding 3 to both sides gives:
3y=3
Therefore, by dividing both sides by 3, y=1.
So the solution is:
x=1, y=1
參考: My brain
👍 0 👎 0
2008-06-16 3:08 pm
ok well you have to eliminate the x and then divide the y so..

the 4x will transfer over to the other side and make 3y=-4x+7

then u must divide the y so dive all by 3 and you get

y=-4/3x+3.333

and do the same on the other
👍 0 👎 0
2008-06-16 3:05 pm
do u mean 2x+3y-5=0?

ok u can take the first eqn and subtract the 2nd eqn. This is to get rid of the y variable.

What u get is eqn(1) - eqn(2): 2x + 5 = 7

from there u can find x, which is 1.
Substitute this into the first eqn again and then u can find y, which is also 1. :)
👍 0 👎 0


收錄日期: 2021-05-01 10:36:53
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