Use substitution method to solve the linear system.
Can someone please check if I did this problem right? Thanks
X+y=12
X+3/2y=3/2
Y=21
x=-9
To check, all you have to do is to see if your solution satisfies both equations.
Your solution satisfies the first, but not the second, so there is an erro
from the first: y = 12-x
first I would multiply the second by 2 to get rid of fractions
2x + 3y = 3
now sub in the first
2x + 3(12-x) = 3
2x + 36 - 3x = 3
-x = -33
x = 33
then y = 12-33 = -21
x = 33 , y = -21
check:
in first:
LS = x+y = 33-21 = 12 = RS
the 2nd:
LS = x + (3/2)y = 33 + (3/2)(-21)
= 33 - 31.5 = 1.5 = 3/2 = RS
To use the substitution method to solve the linear system, we need to isolate one of the variables in one of the equations and substitute it into the other equation.
Given the system of equations:
1) X + y = 12
2) X + (3/2)y = 3/2
Let's solve this system using the substitution method:
From equation (1), we can isolate X:
X = 12 - y
Now, substitute this value for X in equation (2):
12 - y + (3/2)y = 3/2
To eliminate the fraction, we can multiply the entire equation by 2:
2(12 - y) + 3y = 3
Expanding and simplifying:
24 - 2y + 3y = 3
24 + y = 3
Solve for y:
y = 3 - 24
y = -21
Now that we have the value of y, substitute it back into equation (1) to find x:
X + (-21) = 12
X - 21 = 12
X = 12 + 21
X = 33
So, the solution to the given system of equations is:
X = 33
y = -21
From your initial response, it seems you made a mistake. The correct solution to the system is:
X = 33
y = -21