Respuesta :
4b + 5r = 32
b + 8r = 35.....multiply by -4
----------------
4b + 5r = 32
-4b - 32r = - 140 (result of multiplying by -4)
---------------add
- 27r = - 108
r = -108/-27
r = 4 <== each red block weighs 4 oz
b + 8r = 35
b + 8(4) = 35
b + 32 = 35
b = 35 - 32
b = 3 <=== each blue block weighs 3 oz
so 1 red block + 1 blue block weight (3 + 4) = 7 oz <==
b + 8r = 35.....multiply by -4
----------------
4b + 5r = 32
-4b - 32r = - 140 (result of multiplying by -4)
---------------add
- 27r = - 108
r = -108/-27
r = 4 <== each red block weighs 4 oz
b + 8r = 35
b + 8(4) = 35
b + 32 = 35
b = 35 - 32
b = 3 <=== each blue block weighs 3 oz
so 1 red block + 1 blue block weight (3 + 4) = 7 oz <==
following are calculations of the weight:
Given:
blue =4
red= 5
total weight= 32
and
blue =1
red= 8
total weight= 35
To find:
weight of 1 red and 1 blue=?
Solution:
blue =4
red= 5
total weight= 32
So equation:
[tex]\to 4b + 5r = 32...........................(a)[/tex]
similarly
blue =1
red= 8
total weight= 35
So equation:
[tex]\to 1b+8r=35..............(b)[/tex]
Solving the equation (b) and putting its value into equation (a):
[tex]\to 1b+8r=35\\\\\to b= 35-8r\\[/tex]
putting the value into equation (a):
[tex]\leq \to 4(35-8r) + 5r = 32\\\\ \to 140 -32r + 5r = 32\\\\ \to -27r = 32-140\\\\\to -27r=-108\\\\\to r=\frac{108}{27}\\\\\to r=4[/tex]
Putting the value of r in the equation (a):
[tex]\to 4b + 5(4) = 32\\\\\to 4b + 20 = 32\\\\\to 4b= 32-20\\\\\to 4b=12\\\\\to b=\frac{12}{4}\\\\\to b= 3\\[/tex]
Following are the calculation of the weight of
[tex]\to \text{W=1 red block + 1 blue block }\\\\[/tex]
[tex]= 3 + 4\\\\ = 7\\\\[/tex]
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brainly.com/question/9801213