Marthaniceidn
Answered

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-10x+5y=5
x+y=15 substitution and elimination method

Sagot :

Shinalcantaraidn
Using Substitution Method:
-10x + 5y = 5      ----equation 1
x + y = 15    ----equation 2
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From equation 2
x + y = 15
x = 15 - y    -----equation 2'
-------------------
Substitute equation 2' to equation 1
-10x + 5y = 5
-10(15-y) + 5y = 5
Distribute -10
-150 + 10y + 5y = 5
10y + 5y = 5 + 150
15y = 155
y = 31/3
or 
y = 10.33
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Substitute y=31/3 to equation 2'
x = 15 - y
x = 15 - 31/3
x = 14/3
or 
x = 4.67
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Using Elimination Method:
-10x + 5y = 5     ----equation 1
x + y = 15      ----equation 2
-------------------
Multiply equation 2 with 5
 [x + y = 15]    x5
5x + 5y = 75    ------equation 2'
-------------------
Subtract equation 1 from equation 2'
       5x + 5y = 75
-  -10x + 5y = 5
     15x   = 70
         x = 14/3
-------------------
Multiply equation 2 with 10
[x + y = 15]   x10
10x + 10y = 150     ----equation 3
-----------------
Add equations 3 and 1
    10x + 10y = 150
+ -10x + 5y = 5
               15y = 155
                   y = 31/3