Systems of linear equations in three variables?
1st system: x+y+8z = 1 x+y+6z = -1 x+6y+9z = -18 2nd system: 3x-6y+3z = 3 x+3y-z = 0 6x-y-z = 22 3rd system: You Throw a ball straight up from a rooftop. The ball misses the rooftop on its way down and eventually strikes the ground. A mathematical model can be used to describe the relationship for the ball's height above ground, y, after x seconds. X - Y 1 339 3 321 4 264 a. find the quadratic function whose passes through the given points b. use the function in part (a) to find the value for y when x=5
Systems of Linear Equations in three variables?
2x + y - z = 5 -5x - 3y + 2z = 7 x + 4y - 3z = 0 The best way to solve this is to solve for one variable in terms of the others in one equation, and then plug this value into the other two equations. Let's pick the third equation. x + 4y - 3z = 0 implies x = -4y + 3z Plugging this into the other two equations, 2x + y - z = 5 and -5x - 3y + 2z = 7, we have 2(-4y + 3z) + y - z = 5 -5(-4y + 3z) - 3y + 2z = 7 Simplifying both equations, we have -8y + 6z + y - z = 5 20y - 15z - 3y + 2z = 7 And then grouping like terms, -7y + 5z = 5 17y - 13z = 7 Note that at this point, we have a two equations, two unknowns problem. We can now solve this by elimination or substitution; I'll choose substitution. -7y + 5z = 5 implies 5z = 7y + 5, so z = (1/5)(7y + 5) Plugging this into the second equation, 17y - 13z = 7, we have 17y - 13[(1/5)(7y + 5)] = 7 17y - (13/5)(7y + 5) = 7 Multiplying both sides by 5 to get rid of the fraction, 85y - 13(7y + 5) = 7 85y - 91y - 65 = 7 -6y = 72 y = -12 Now that we have y = -12, we can get z. z = (1/5)(7y + 5) = (1/5) (7(-12) + 5) z = (1/5) (-84 + 5) z = (1/5) (-79) z = -79/5 Now that we have y and z, we can get x.
How do you solve system of linear equations in three variables?
pick the first 2 of the equations and use the elimination method to get another equations with only 2 variables, such as the 3rd equation. you do this by eliminating the variable that is not present in all of the equations, 4x -2y + 2z = -8 (every term multiplied by 2) 3x + y - 2z = 0 ______________ ( add the two equations) 7x-y=-8 (z is gone) now use the two equations to figure out x and y 7x-y=-8 -3x + y = 4 (every term multiplyed by -1) ________ ( add the two equations) 4x=-4 x=-1 3(-1) - y = -4 -y= -4+3 y=1 now use one of the equations with the z variable and solve 2(-1) - (1) + z = -4 z= -4 +2 +1 z=-1 therefore the point of intersection is (-1,1,-1) all the other answerers used the substitution method, which works, but is more complicated and slower than the elimination method, besides, this way is easier to learn and less confusion 8D
Solve systems of linear equations in three variables?
Make any equation in to a single variable = other 2 variables. Let take the 2nd one, we can say -4y = 8y + 8z -8, y = -2x -2z + 2 So plug that into the other 2 equations a) 16x -8(-2x - 2z +2) + 4z = -2 16x +16x +16z -16 +4z = -2 32x + 20z =14 b) -12x - 4(-2x - 2z +2) -16z = -7 -12x +8x +8z -8 -16z = -7 -4x -8z = 1 32x + 20z = 14 -4x - 8z = 1 Now we have 2 linear equation, with 2 variables, and can solve whatever method you choose
Algebra II question- linear equation with three variables word problem?
If you want x, y, and z x=6, y=2, and z=6 3 equations x=z 4x+6y+5z=66 x+y+z=14
Solving Systems of Linear Equations in Three Variables!?
since it is multiple choice you do not need to solve it. You just have to take the answers and sub it into the equations. If it works for all equations then it is a solution to the system. 1st questions try subbing (-4,0,2) -4 -0 +2*2 = -10 0=10 This is not true, so this is not a solution to the system (-4,2,0) -4 -2 +2*0= -10 -6 = 10 NOT TRUE thus not a solution And (0,0,0) give 0=-10 which is also not true Solution must be (0,2,-4) 2. Answer is (-4,4,-1) You can sub this into each equation to check. 3. The last one has no solution. Two of the lines are parallel. If there is a D choice, take it. There is no solution.
Solving systems of linear equations in three variables?
Add #1 and #2 to eliminate X (X + -X) + (2Y+Y) + (-Z+3Z) = (3+ -5) 3Y+2Z =-2 Now add 3 times #2 to #3 to eliminate X (-3X + 3X) + (3Y + Y) + (9Z + 2Z) = (-15+4) 4Y + 11Z = -11 Now you have two equations with two unknowns 3Y+2Z=-2 4Y+11Z=-11 You can eliminate Y by multiplying by -4 for the first and multiplying by 3 to the second and then adding them together (-12Y+12Y) + (-8Z+33Z) = (8+-33) 25Z = -25 Z = -1 Now, plug Z back into the above equation 3Y+2Z=-2 3Y+2(-1) = -2 3Y-2=-2 3Y=0 Y=0 Now plug Y and Z back into #1 X+2Y-Z=3 X+0+1=3 X=2
How do you solve linear equations with three variables?
Doing it with matrices is the easiest. If you have a Ti-83+ or a Ti-84+, you can do this: hit 2nd + Matrix (the x^-1 key) Go to Edit Hit enter make your matrix a 3x4 matrix plug in your values: [3 1 1 9] [1 1 1 5] [1 2 1 9] hit 2nd + Quit (mode) Go back to the matrix menu go to math go to B: rref( and hit enter go back to matrix menu hit enter hit enter again and your answers are the far right column: u = 2 v = 4 w = -1
Help with solving systems of equations in three variables?
Is there any way to do these type of problems on a TI-84 Plus Calculator. Ex. 4x + 4y - 2z = 3 -6x - 6y + 6z = 5 2x - 3y - 4z = 2 Thanks