solve for x, y and z when
2x + 4y + 2z = 144
4x + y + 0.5z = 120
x + 3y + 4z = 144
Multiply Eq2 by -4 and add Eq1 and Eq2:
2x + 4y + 2z = 144
-16x - 4y - 2z = -480
Sum: -14x = -336
X = 24
Multiply Eq1 by -2; add Eq1 and Eq3:
-4x - 8y - 4z = -288
+x + 3y + 4z = 144
Sum: -3x - 5y = -144.
X = 24. Solve for y:
-3*24 - 5y = -144
-72 - 5y = -144
-5y = -144 + 72 = -72
Y = 14.4
Eq1: 2x + 4y + 2z = 144
X = 24, Y = 14.4. Solve for Z.
2*24 + 4*14.4 + 2z = 144
48 + 57.6 + 2z = 144
2z = 144 - 48 - 57.6 = 38.4
Z = 19.2
To solve for x, y, and z, we can use the method of substitution or elimination. Here, we will use the method of elimination.
Step 1: Multiply the equations by suitable constants to make the coefficients of x, y, or z the same in two equations, allowing us to eliminate a variable.
Let's start by multiplying the second equation by 2 to eliminate the x coefficient:
2 * (4x + y + 0.5z) = 2 * 120
8x + 2y + z = 240 ---- (Equation 4)
Step 2: Now, let's eliminate the z variable.
Multiply the first equation by 2 and subtract it from the third equation:
2*(2x + 4y + 2z) - (x + 3y + 4z) = 2*144 - 144
4x + 8y + 4z - x - 3y - 4z = 288 - 144
3x + 5y = 144 ---- (Equation 5)
Step 3: We now have two equations (Equation 4 and Equation 5) involving x and y. We will eliminate y from these equations.
Multiply Equation 4 by 5 and Equation 5 by 2:
5*(8x + 2y + z) = 5 * 240
40x + 10y + 5z = 1200 ---- (Equation 6)
2*(3x + 5y) = 2 * 144
6x + 10y = 288 ---- (Equation 7)
Step 4: Subtract Equation 7 from Equation 6 to eliminate y:
(40x + 10y + 5z) - (6x + 10y) = 1200 - 288
34x + 5z = 912 ---- (Equation 8)
Step 5: Now, we can solve the system of equations involving x and z.
From Equation 8, solve for z in terms of x:
34x + 5z = 912
5z = 912 - 34x
z = (912 - 34x)/5 ---- (Equation 9)
Step 6: We can substitute Equation 9 into Equation 8 and solve for x:
34x + 5((912 - 34x)/5) = 912
34x + 912 - 34x = 912
912 = 912
Step 7: Since the equation is true, it means any value of x will satisfy the equation. Let's choose x = 0.
Substituting x = 0 into Equation 9, we can solve for z:
z = (912 - 34(0))/5
z = 912/5
z = 182
Step 8: Now that we have x = 0 and z = 182, we can substitute these values into Equation 5 to solve for y:
3x + 5y = 144
3(0) + 5y = 144
5y = 144
y = 144/5
y = 28.8
Therefore, the solution to the system of equations is:
x = 0
y = 28.8
z = 182
To solve for x, y, and z in the given system of equations, we can use the method of substitution or elimination. Here, we'll use the elimination method:
1. Multiply the first equation by 2, the second equation by 4, and the third equation by 1 to make the coefficient of x in each equation match:
Equation 1: 4x + 8y + 4z = 288
Equation 2: 4x + y + 0.5z = 120
Equation 3: x + 3y + 4z = 144
2. Subtract Equation 2 from Equation 1 to eliminate x:
(4x - 4x) + (8y - y) + (4z - 0.5z) = (288 - 120)
7y + 3.5z = 168
3. Subtract Equation 3 from Equation 2 to eliminate x:
(4x - x) + (y - 3y) + (0.5z - 4z) = (120 - 144)
3x - 2y - 3.5z = -24
4. Now, we have a system of two equations in variables y and z:
Eq1: 7y + 3.5z = 168
Eq2: 3x - 2y - 3.5z = -24
5. Multiply Equation 2 by 2 to make the coefficient of y the same magnitude as in Equation 1:
6x - 4y - 7z = -48
6. Add Equation 1 and Equation 2 to eliminate y:
(7y - 4y) + (3.5z - 7z) = (168 - 48)
3y - 3.5z = 120
7. We now have a new system of two equations:
Eq1: 3y - 3.5z = 120
Eq2: 6x - 4y - 7z = -48
8. We can now solve this system of equations. For simplicity, we'll solve for y in terms of z using Equation 1:
3y = 120 + 3.5z
y = (120 + 3.5z) / 3
9. Substitute this expression for y in Equation 2:
6x - 4((120 + 3.5z) / 3) - 7z = -48
10. Simplify Equation 2 and solve for x:
6x - (480 + 14z) / 3 - 7z = -48
Multiply through by 3 to clear the fraction:
18x - 480 - 14z - 21z = -144
18x - 35z = 336
18x = 336 + 35z
x = (336 + 35z) / 18
11. Now substitute the expressions for x and y into Equation 3:
(336 + 35z) / 18 + 3((120 + 3.5z) / 3) + 4z = 144
12. Simplify Equation 3 and solve for z:
(336 + 35z) / 18 + (120 + 3.5z) + 4z = 144
(336 + 35z) / 18 + 120 + 3.5z + 4z = 144
Multiply through by 18 to clear the fraction:
336 + 35z + 18(120 + 3.5z + 4z) = 144 * 18
336 + 35z + 2160 + 63z + 72z = 2592
170z = 2592 - 336 - 2160
170z = 96
13. Solve for z:
z = 96 / 170
z ≈ 0.5647
14. Substitute the value of z back into the expressions for x and y:
x = (336 + 35(0.5647)) / 18 ≈ 7.1765
y = (120 + 3.5(0.5647)) / 3 ≈ 20.6471
Hence, the solution to the system of equations is approximately:
x ≈ 7.1765
y ≈ 20.6471
z ≈ 0.5647