To solve the given system of equations, we can use the method of substitution.
First, let's solve the first equation for x in terms of y and z:
3x + 2y + z = 7
3x = 7 - 2y - z
x = (7 - 2y - z)/3
Now substitute this value of x into the second equation:
5x + 5y + 4z = 3
5((7 - 2y - z)/3) + 5y + 4z = 3
(35 - 10y - 5z)/3 + 5y + 4z = 3
35 - 10y - 5z + 15y + 12z = 9 (Multiplying both sides by 3)
-10y + 15y - 5z +12z = 9-35
5y + 7z = -26 (Simplifying the equation)
Now we have two equations:
Equation 1: x = (7 - 2y - z)/3
Equation 2: 5y + 7z = -26
We can solve this system by substitution:
Substitute the value of x from Equation 1 into Equation 2:
5y + 7z = -26
5((7 - 2y - z)/3) + 7z = -26
(35 - 10y - 5z)/3 + 7z = -26
(35 - 10y - 5z + 21z)/3 = -26
(35 - 10y + 16z)/3 = -26
35 - 10y + 16z = -78 (Multiplying both sides by 3)
-10y + 16z = -113 (Simplifying the equation)
Now we have two equations:
Equation 1: 5y + 7z = -26
Equation 2: -10y + 16z = -113
We can solve this system using any method, such as substitution or elimination. Let's use elimination to eliminate y:
Multiply Equation 1 by 2 and Equation 2 by 5 to make the coefficients of y in both equations opposite:
2(5y + 7z) = 2(-26)
5(-10y + 16z) = 5(-113)
10y + 14z = -52
-50y + 80z = -565
Now add these equations to eliminate y:
(10y + 14z) + (-50y + 80z) = -52 + (-565)
10y - 50y + 14z + 80z = -52 - 565
-40y + 94z = -617
We now have a new equation:
Equation 3: -40y + 94z = -617
Now we have two equations:
Equation 3: -40y + 94z = -617
Equation 2: -10y + 16z = -113
We can use elimination again to eliminate y:
Multiply Equation 3 by 1/4 to make the coefficients of y in both equations opposite:
(1/4)(-40y + 94z) = (1/4)(-617)
-10y + 23.5z = -154.25
Now add these equations to eliminate y:
(-10y + 23.5z) + (-10y + 16z) = -154.25 + (-113)
-10y - 10y + 23.5z + 16z = -154.25 - 113
-20y + 39.5z = -267.25
We now have a new equation:
Equation 4: -20y + 39.5z = -267.25
Now we have two equations:
Equation 4: -20y + 39.5z = -267.25
Equation 2: -10y + 16z = -113
We can use elimination again to eliminate y:
Multiply Equation 4 by 1/2 to make the coefficients of y in both equations opposite:
(1/2)(-20y + 39.5z) = (1/2)(-267.25)
-10y + 19.75z = -133.625
Now add these equations to eliminate y:
(-10y + 19.75z) + (-10y + 16z) = -133.625 + (-113)
-10y - 10y + 19.75z + 16z = -133.625 - 113
-20y + 35.75z = -246.625
We now have a new equation:
Equation 5: -20y + 35.75z = -246.625
Now we have two equations:
Equation 5: -20y + 35.75z = -246.625
Equation 2: -10y + 16z = -113
We can use elimination again to eliminate y:
Multiply Equation 5 by 1/2 to make the coefficients of y in both equations opposite:
(1/2)(-20y + 35.75z) = (1/2)(-246.625)
-10y + 17.875z = -123.3125
Now add these equations to eliminate y:
(-10y + 17.875z) + (-10y + 16z) = -123.3125 + (-113)
-10y - 10y + 17.875z + 16z = -123.3125 - 113
-20y + 33.875z = -236.3125
We now have a new equation:
Equation 6: -20y + 33.875z = -236.3125
Now we have two equations:
Equation 6: -20y + 33.875z = -236.3125
Equation 2: -10y + 16z = -113
Let's solve Equation 2 for y:
-10y + 16z = -113
-10y = -113 - 16z
y = (113 + 16z)/10
y = (113/10) + (16/10)z
y = 11.3 + 1.6z
Now substitute this value of y into Equation 6:
-20(11.3 + 1.6z) + 33.875z = -236.3125
-226 - 32z + 33.875z = -236.3125
1.875z = -236.3125 + 226
1.875z = -10.3125
z = -10.3125/1.875
z = -5.5
Now substitute this value of z into Equation 2 to solve for y:
-10y + 16(-5.5) = -113
-10y - 88 = -113
-10y = -113 + 88
-10y = -25
y = -25/(-10)
y = 2.5
Finally, substitute the values of y and z into Equation 1 to solve for x:
x = (7 - 2(2.5) - (-5.5))/3
x = (7 - 5 - (-5.5))/3
x = (7 - 5 + 5.5)/3
x = 7.5/3
x = 2.5
Therefore, the solution to the system of equations is x = 2.5, y = 2.5, and z = -5.5.
Solve the system
3x+2y+z=7
5x+5y+4z=3
1 answer
1 answer