Can you help with this equation? Thanks. Solve for x and y and z for the following. x plus y plus z equals 9 x plus 2y plus 3z equals 23 x plus 5y plus -3z equals -7
-4(7x-3y-z)-(-6)(9x-z+2)
The solution is (x, y, z) = (2.5, 1, 2.5).
10x + 2y + 8z - 5x + 4z - 4y = 10x - 5x + 2y - 4y + 8z + 4z = (10 - 5)x + (2 - 4)y + (8 + 4)z = 5x - 2y + 12z
You can't simplify it any further than 3x+2y+6z=24, unless you have been given values of x,y and z or other equations to find x,y and z.
(x+y)^2+z^2=x^2+y^2+z^2+2xy or ((x+y)^2+z)^2= (x^2+y^2+2xy+z)^2= x^4+y^4+z^2+6x^2y^2+4x^3y+2x^2z^2+4xy^3+4xyz^2+2z^2y^2
True
you can not solve this equation
x+2y-6=z -z -z x+2y-z-6=0 +6 +6 ---------> x+2y-z=6 3y-2z=7 ---------> 0x+3y-2z=7 4+3x=2y-5z -3x -3x ---------> -3x+2y-5z=4 Put them into a matrix, for x,y,z and their answers. Solve for [A]-1[B], and the answer comes to: x= 1.75, y= 1.5, and z= -1.25
Can you help with this equation? Thanks. Solve for x and y and z for the following. x plus y plus z equals 9 x plus 2y plus 3z equals 23 x plus 5y plus -3z equals -7
x-(x-2y)+z=-9 3y-2z=4 2x+y+5z=-5 x-(x-2y)+z=-9 --distribute negative through bracket--> x-x+2y+z=-9 x-x+2y+z=-9 --combine like terms (x)--> 2y+z=-9 2y+z=-9 --isolate z--> z=-9-2y substitute z into second equation: 3y-2(-9-2y)=4 --multiply -2 through bracket--> 3y+18+4y=4 3y+18+4y=4 --subtract 18 from both sides--> 3y+4y=-14 3y+4y=-14 --combine like terms--> 7y=-14 7y=-14 --divide both sides by 7--> y=-2 Substitute value of y into simplified first equation: z=-9-2(-2) --multiply through the bracket--> z=-9+4 z=-9+4 --combine like terms--> z=-5 Substitute values of y and z into third equation: 2x+(-2)+5(-5)=-5 --multiply through brackets--> 2x-2-25=-5 2x-2-25=-5 --combine like terms--> 2x-27=-5 2x-27=-5 --add 27 to both sides--> 2x=22 2x=22 --divide both sides by 2--> x=11 x=11, y=-2, z=-5
-4(7x-3y-z)-(-6)(9x-z+2)
The solution is (x, y, z) = (2.5, 1, 2.5).
10x + 2y + 8z - 5x + 4z - 4y = 10x - 5x + 2y - 4y + 8z + 4z = (10 - 5)x + (2 - 4)y + (8 + 4)z = 5x - 2y + 12z
x = -1, y = 2, z = -3.x - y - z = 03x + y - z = 2x + 2y + z = 0Adding equation 3 to equations 1 and 2 gives new equations 1 & 2:2x + y = 04x + 3y = 2Doubling the [new] equation 1 and subtracting from the [new] equation 2 gives:y = 2Substituting back into [new] equation 1 gives:2x + 2 = 0 ==> x = -1Substituting back into the original equation 1 gives:-1 - 2 - z = 0 ==> z = -3Check by substituting back into original equations:3x + y - z = 3(-1) + (2) - (-3) = -3 + 2 + 3 = 2x + 2y + z = (-1) + 2(2) + (-3) = -1 + 4 - 3 = 0
You can't simplify it any further than 3x+2y+6z=24, unless you have been given values of x,y and z or other equations to find x,y and z.
If th equestion meant: (x+y+z)^2The expansion is:(x+y+z)^2= x^2+2xy+y^2+2yz+z^2+2zx