SOLUTION: solve the system of equations 3y + 2z =12 and y

SOLUTION: solve the system of equations 3y + 2z =12 and y - z=9

Question 320956: solve the system of equations 3y + 2z =12 and y - z=9
Answer by Sunny Day(15) (Show Source): You can put this solution on YOUR website!
3y + 2z = 12 --------(1)
y - z = 9 --------(2)
Solving this means finding the values of 'y' and 'z'. This can be done by eliminating one of the variables from the given equations. Here, since neither of the variables has the same coefficient in the two equations, we need to make the coefficients of one the variables the same in both the equations.
Let us make the coefficient 'z' the same in both the equations. This can be done by multiplying the first equation with the coefficient of 'z' (you need to take the absolute value only) in the second and the second equation with the coefficient of 'z' in the first.
so multiplying (1) with 1 ----> 1(3y + 2z) = 1 X 12 ----> 3y + 2z = 12 ---(3)
& multiplying (2) with 2 ------> 2(y - z) = 2 X 9 -----> 2y - 2z = 18 ---(4)
_________________
Now we will use (3) and (4) for elimination which can be done by just adding the two.
SO (3) + (4) -----> 3y + 2z + 2y -2z = 12 + 18
5y + 0 = 30 (SINCE 2Z - 2Z = 0)
5y = 30
y = 30/5 = 6
Now put this value of y in (1),(2),(3) or (4) to get 'z'
putting in (2), 6 - z = 9
-z = 9 - 6 = 3
z = -3.
SO the solution is y = 6 and z = -3.
Cross-checking
Put these values in (1).
ie. 3X6 + 2X-3 = 18 + -6 = 18 - 6 = 12, which is the RHS of (2).
SO the solution is verified.

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