Solving Linear Equations and Inequalities
Solve by eliminating x (this is solving 3 linear equations) then substitute to the other two equation
x + y + 5z =2 (1)
4x - 3y + 5z =3 (2)
3x - 2y + 5z=1 (3)
my teacher gave me the answer she said the solution set for this should be {(7,5-2)}
pleeeeeeeeeeeeeaaaseee :(
http://www.jiskha.com/display.cgi?id=1461567386#1461567386.1461580442
Why eliminate x when the obvious choice would be to eliminate z ??
#2 - #1 ---> 3x - 4y = 1 **
#2 - #3 ---> x - y = 2 or x = y+2
sub that into **
3(y+2) - 4y = 1
3y + 6 - 4y = 1
-y = -5
y = 5 , then
x = 7
back into #1:
7 + 5 + 5z = 2
5z = -10
z = -2
so the solution is (7,5,-2)
Depending on the coefficients in the equations, it is often easier to use a combination of methods, rather than one rigid approach
xis what needs to be eliminated. . . my teacher already try eliminated the z
Ok,
multiply the first by 4, then subtract the 2nd
multiply the first by 3, then subtract the third.
Now you have two equations with y's and z's
carry on
To solve the system of equations using the elimination method, you need to eliminate one variable at a time. Let's start with eliminating the variable x.
First, let's eliminate x from equations (2) and (3). Multiply equation (1) by 4 and equation (3) by 3 to make the coefficients of x in equations (2) and (3) the same as the coefficient of x in equation (1).
Multiplying equation (1) by 4:
4x + 4y + 20z = 8 (4)
Multiplying equation (3) by 3:
9x - 6y + 15z = 3 (5)
Now, subtracting equation (5) from equation (4) will eliminate x:
(4x + 4y + 20z) - (9x - 6y + 15z) = 8 - 3
Simplifying the equation:
-5x + 10y + 5z = 5 (6)
Now, let's eliminate x from equation (1) and (6). Multiply equation (1) by 5 and equation (6) by 1 to make the coefficients of x in both equations the same.
Multiplying equation (1) by 5:
5x + 5y + 25z = 10 (7)
Multiplying equation (6) by 1:
-5x + 10y + 5z = 5 (8)
Adding equation (7) and equation (8) will eliminate x:
(5x + 5y + 25z) + (-5x + 10y + 5z) = 10 + 5
Simplifying the equation:
15y + 30z = 15 (9)
Now, we have reduced the system of equations from three equations to two equations. Let's solve them:
From equation (9), we can isolate y by dividing both sides by 15:
15y + 30z = 15
y + 2z = 1
Now, we can substitute this value of y into equation (6) to solve for z:
-5x + 10(y + 2z) + 5z = 5
-5x + 10y + 20z + 5z = 5
2z = 2
z = 1
Now that we have the value of z, we can substitute it back into equation (7) or (8) to solve for y:
From equation (7):
5x + 5y + 25z = 10
5x + 5y + 25(1) = 10
5x + 5y + 25 = 10
5x + 5y = -15
x + y = -3
x = -y - 3
Let's substitute this relationship between x and y into equation (1):
(-y - 3) + y + 5(1) = 2
-y - 3 + y + 5 = 2
2 = 2
Since 2 = 2 is a true statement, we can conclude that this system of equations has infinitely many solutions. Therefore, it does not have a unique solution. Instead of a solution set, we have a solution relationship x = -y - 3.
It seems there might be an error in the answer you provided from your teacher. The correct solution set cannot be determined without more information.