Could you help me to solve the following simulatenous equation 15x- 19y = 22, 20x - 23y = 34?
Could you help me to solve the following simulaneous equation
15x - 19y = 22, 20x - 23y 34
If you could give me explanation's as you go that would be very helpful
Answers:
15x - 19y = 22,
Becomes
15x = 22 +19y
x = (22 + 19y)/15
So if you stick it into:
20x - 23y = 34
You get:
20*((22 + 19y)/15) - 23y = 34
Which goes to:
(440 + 380y)/15 -23y = 34
Then:
Seperate it out to:
440/15 + 380y/15 -23y = 34
Then to:
29.333 +25.333y -23y = 34
2.333y = 34 - 29.333 = 4.666
y = 4.666/2.333 = 2
Then taking:
15x - 19y = 22
15x - 19*2 = 22
15x - 38 = 22
15x = 22 + 38 = 60
x = 60/15 = 4
x = 4
y = 2.
Since you have a system of 2 equations, solve one for a single variable, then input that value into the other. Then solve that for the other variable and input that into the first euqation and solve for the first variable.
you have to multiply one whole equation or both so that they will cancel out one of the variables.i would multiply the first one by 4 and the second one by 3 in order to canel out the x variables.
(15x - 19y = 22)4 = 60x - 76y = 88
(20x - 23y = 34)3 = 60x - 69y = 102
then subtract one from the other this will give you
-7y = -14
and y = 2
then plug this y value back into an original equation to get the x
20x - 23(2) = 34
20x - 46 = 34
20x = 80
x = 4
those are your two answers.
multiply first eqn by 4 and 2nd eqn by 3 to make 60x in both
60x-76y=88
60x-69y=102 so 7y=14 by subtracting upper equation from lower one so y=2
so15x-38=22 so 15x=60 so x=4
Do your own homework!! It is ok to ask one for the theory, but it is obvious posting 4 different questions u just need your homework doing!
15x-19y=22...(1)
20x-23y=34...(2)
5x-4y =12...(2)-(1)
15x-12y=36...mult by 3
15x-19y=22...(1)
>>>>7y=14 >>>>y=2...by subtraction
>>>>>>>>>>>>>>x=4...by substitution
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15x - 19y = 22, 20x - 23y 34
If you could give me explanation's as you go that would be very helpful
Answers:
15x - 19y = 22,
Becomes
15x = 22 +19y
x = (22 + 19y)/15
So if you stick it into:
20x - 23y = 34
You get:
20*((22 + 19y)/15) - 23y = 34
Which goes to:
(440 + 380y)/15 -23y = 34
Then:
Seperate it out to:
440/15 + 380y/15 -23y = 34
Then to:
29.333 +25.333y -23y = 34
2.333y = 34 - 29.333 = 4.666
y = 4.666/2.333 = 2
Then taking:
15x - 19y = 22
15x - 19*2 = 22
15x - 38 = 22
15x = 22 + 38 = 60
x = 60/15 = 4
x = 4
y = 2.
Since you have a system of 2 equations, solve one for a single variable, then input that value into the other. Then solve that for the other variable and input that into the first euqation and solve for the first variable.
you have to multiply one whole equation or both so that they will cancel out one of the variables.i would multiply the first one by 4 and the second one by 3 in order to canel out the x variables.
(15x - 19y = 22)4 = 60x - 76y = 88
(20x - 23y = 34)3 = 60x - 69y = 102
then subtract one from the other this will give you
-7y = -14
and y = 2
then plug this y value back into an original equation to get the x
20x - 23(2) = 34
20x - 46 = 34
20x = 80
x = 4
those are your two answers.
multiply first eqn by 4 and 2nd eqn by 3 to make 60x in both
60x-76y=88
60x-69y=102 so 7y=14 by subtracting upper equation from lower one so y=2
so15x-38=22 so 15x=60 so x=4
Do your own homework!! It is ok to ask one for the theory, but it is obvious posting 4 different questions u just need your homework doing!
15x-19y=22...(1)
20x-23y=34...(2)
5x-4y =12...(2)-(1)
15x-12y=36...mult by 3
15x-19y=22...(1)
>>>>7y=14 >>>>y=2...by subtraction
>>>>>>>>>>>>>>x=4...by substitution
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