Asked by Alyssa
i need help with these problems (either a solution or how to plug them into a graphing calculator):
1) Solve by substitution or elimination
4/x + 1/y + 2/z = 4
2/x + 3/y - 1/z = 1
1/x + 1/y + 1/z = 4
2)Solve the system of equations by substitution.
x^2 + y^2 = 25
x - y = 7
3) Solve the system of equations by substitution. Round the answer to the nearest hundredths.
9x^2 + 16y^2 = 140
x^2 - 4y^2 = 4
4)Solve by elimination:
1/x + 3/y = 7
4/x - 2/y = 1
Thanks!
1) Solve by substitution or elimination
4/x + 1/y + 2/z = 4
2/x + 3/y - 1/z = 1
1/x + 1/y + 1/z = 4
2)Solve the system of equations by substitution.
x^2 + y^2 = 25
x - y = 7
3) Solve the system of equations by substitution. Round the answer to the nearest hundredths.
9x^2 + 16y^2 = 140
x^2 - 4y^2 = 4
4)Solve by elimination:
1/x + 3/y = 7
4/x - 2/y = 1
Thanks!
Answers
Answered by
Reiny
1. add the 2nd and 3rd
----> 3/x + 4/y = 5 (#4)
double the 2nd and add to the 1st
---> 8/x + 7/y = 6 (#5)
8 times #4 --> 24/x + 32/y = 40
3 times #5 --> 24/x + 21/y = 18
subtract:
11/y = 22
11 = 22y
y = 1/2
sub back into either #4 or #4 to get x
then back into 1st to get z
2. from the second: x = y+7
into the first: (y+7)^2 + y^2 = 25
2y^2 + 14y + 24 = 0
y^2 + 7y + 12 = 0
(y+3)(y+4) = 0
y = -3 or y = -4
sub both of those into x = y+7 to get their corresponding x values
3. from 2nd ---> x^2 = 4y^2 + 4
sub into 1st, etc
4. easier than #1
multiply 1st by 4, then subtract the 2nd
----> 3/x + 4/y = 5 (#4)
double the 2nd and add to the 1st
---> 8/x + 7/y = 6 (#5)
8 times #4 --> 24/x + 32/y = 40
3 times #5 --> 24/x + 21/y = 18
subtract:
11/y = 22
11 = 22y
y = 1/2
sub back into either #4 or #4 to get x
then back into 1st to get z
2. from the second: x = y+7
into the first: (y+7)^2 + y^2 = 25
2y^2 + 14y + 24 = 0
y^2 + 7y + 12 = 0
(y+3)(y+4) = 0
y = -3 or y = -4
sub both of those into x = y+7 to get their corresponding x values
3. from 2nd ---> x^2 = 4y^2 + 4
sub into 1st, etc
4. easier than #1
multiply 1st by 4, then subtract the 2nd
Answered by
LAUGHSTOP
what? im confused
Answered by
Michael omoloye
Okay;
let,(x-y)^2=x^2+y^2-2xy.
But,x-y=7 and x^2+y^2=25.by apply the formula.
(7)^2=25-2xy
49=25-2xy
xy=-12. Then also,
(x+y)^2=(x-y)^2+4xy
(x+y)^2=49-48.
X+y=+/-1. Then solve the eqn.
X+y=+/-1
x-y=7. Then
x=3 or 4.
Y=-3 or -4.
let,(x-y)^2=x^2+y^2-2xy.
But,x-y=7 and x^2+y^2=25.by apply the formula.
(7)^2=25-2xy
49=25-2xy
xy=-12. Then also,
(x+y)^2=(x-y)^2+4xy
(x+y)^2=49-48.
X+y=+/-1. Then solve the eqn.
X+y=+/-1
x-y=7. Then
x=3 or 4.
Y=-3 or -4.
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