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Sagot :
let 'x' be the cost of 1 apple
'y' be the cost of 1 pear
'z' be the cost of 1 orange
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Abe paid Php1.90 for 1 pear and 3 apples
y + 3x = 1.90 ------equation 1
---------------------------------
Joan paid Php1.60 for 1 pear, 1 apple and 1 orange
x + y + z = 1.60 -----equation 2
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Latonya paid Php1.70 for 2 apples and 1 orange
2x + z = 1.70 ----equation 3
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express y and z in terms of x
from equation 1
y + 3x = 1.90
y = -3x + 1.90 ----equation 1'
from equation 3
2x + z = 1.70
z = -2x + 1.70 ----equation 3'
---------------------------------
Substitute equations 1' and 3' to equation 2
x + y + z = 1.60
x + (-3x + 1.90) + (-2x + 1.70) = 1.60
x - 3x + 1.90 - 2x + 1.70 = 1.60
x - 3x - 2x = 1.60 - 1.70 - 1.90
-4x = -2
x = 0.50
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Substitute x=0.50 to equations 1' and 3'
y = -3x + 1.90
y = -3(0.50) + 1.90
y = -1.50 + 1.90
y = 0.40
z = -2x + 1.70
z = -2(0.50) + 1.70
z = -1 + 1.70
z = 0.70
------------------------------
Therefore apple costs Php0.50 each
pear costs Php0.40 each and
orange costs Php0.70 each
'y' be the cost of 1 pear
'z' be the cost of 1 orange
----------------------------------
Abe paid Php1.90 for 1 pear and 3 apples
y + 3x = 1.90 ------equation 1
---------------------------------
Joan paid Php1.60 for 1 pear, 1 apple and 1 orange
x + y + z = 1.60 -----equation 2
-------------------------------
Latonya paid Php1.70 for 2 apples and 1 orange
2x + z = 1.70 ----equation 3
---------------------------------
express y and z in terms of x
from equation 1
y + 3x = 1.90
y = -3x + 1.90 ----equation 1'
from equation 3
2x + z = 1.70
z = -2x + 1.70 ----equation 3'
---------------------------------
Substitute equations 1' and 3' to equation 2
x + y + z = 1.60
x + (-3x + 1.90) + (-2x + 1.70) = 1.60
x - 3x + 1.90 - 2x + 1.70 = 1.60
x - 3x - 2x = 1.60 - 1.70 - 1.90
-4x = -2
x = 0.50
------------------------------
Substitute x=0.50 to equations 1' and 3'
y = -3x + 1.90
y = -3(0.50) + 1.90
y = -1.50 + 1.90
y = 0.40
z = -2x + 1.70
z = -2(0.50) + 1.70
z = -1 + 1.70
z = 0.70
------------------------------
Therefore apple costs Php0.50 each
pear costs Php0.40 each and
orange costs Php0.70 each
let:
x-pear
y-apples
z-orange
x+3y=1.9
x+y+z=1.6
2x+z=1.7
Lets find the value of x in the first equation
x=1.9-3y
Lets find the value of z in the third equation.
z=1.7-2x
z=1.7-3.8+6y
z=-2.1+6y
Substitute the value on the 2nd equation.
1.9-3y+y-2.1+6y=1.6
4y-0.2=1.6
4y=1.8
y=0.45
Substitute the value of x:
x=1.9-3(0.45)
x=1.9-1.35
x=0.55
Substitute the value of x
z=1.7-2(0.55)
z=1.7-1.1
z=0.6
Hope this helps =)
x-pear
y-apples
z-orange
x+3y=1.9
x+y+z=1.6
2x+z=1.7
Lets find the value of x in the first equation
x=1.9-3y
Lets find the value of z in the third equation.
z=1.7-2x
z=1.7-3.8+6y
z=-2.1+6y
Substitute the value on the 2nd equation.
1.9-3y+y-2.1+6y=1.6
4y-0.2=1.6
4y=1.8
y=0.45
Substitute the value of x:
x=1.9-3(0.45)
x=1.9-1.35
x=0.55
Substitute the value of x
z=1.7-2(0.55)
z=1.7-1.1
z=0.6
Hope this helps =)
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