solve the system of elimination. {×+5y-z=-10; 2×-y+5z+5z=-9; 2×-10y-5z=0
To solve the system of equations by elimination, we will first eliminate the variable ×.
The system of equations is:
1. × + 5y - z = -10
2. 2× - y + 5z + 5z = -9
3. 2× - 10y - 5z = 0
We can solve the system of equations in the following steps:
Equation 2 can be simplified by adding the two terms with z:
2× - y + 10z = -9
Now we have two equations with the same value of ×:
× + 5y - z = -10
2× - y + 10z = -9
To eliminate ×, we need to multiply equation 1 by -2:
-2(× + 5y - z) = -2(-10)
-2× - 10y + 2z = 20
Now we have:
-2× - 10y + 2z = 20
2× - y + 10z = -9
Adding the equations together, the × variable is eliminated:
(-2× + 2×) + (-10y - y) + (2z + 10z) = 20 - 9
-11y + 12z = 11
Now we have the simplified equation:
-11y + 12z = 11 ...(4)
Next, we need to eliminate the variable × again in the third equation:
2× - 10y - 5z = 0
Multiplying the first equation by 2, we get:
2(× + 5y - z) = 2(-10)
2× + 10y - 2z = -20
Now we have:
2× + 10y - 2z = -20
2× - 10y - 5z = 0
Adding the equations together, the × variable is eliminated:
(2× + 2×) + (10y - 10y) + (-2z - 5z) = -20 + 0
0 + (-7z) = -20
-7z = -20
Simplifying the equation, we find:
z = -20 / -7
z = 20/7 ...(5)
Substituting the value of z in equation 4, we find:
-11y + 12(20/7) = 11
-11y + 240/7 = 11
Multiplying the equation by 7 to eliminate the fraction, we get:
-77y + 240 = 77
-77y = 77 - 240
-77y = -163
y = -163 / -77
y = 163/77 ...(6)
Finally, substituting the values of y and z in equation 1, we find:
x + 5(163/77) - 20/7 = -10
Multiplying the equation by 77 to eliminate the fraction, we get:
77x + 5(163) - 20(11) = -770
77x + 815 - 220 = -770
77x = -770 - 815 + 220
77x = -1365
x = -1365 / 77
x = -45/11 ...(7)
Therefore, the solution to the system of elimination is:
x = -45/11, y = 163/77, z = 20/7.
To solve this system of equations using the elimination method, we will eliminate one variable at a time.
Step 1: Let's eliminate the variable x from the equations:
Equation 1: x + 5y - z = -10
Equation 2: 2x - y + 5z + 5z = -9
To eliminate x, we'll multiply Equation 1 by 2:
2(x + 5y - z) = 2(-10)
2x + 10y - 2z = -20
Now we can subtract Equation 2 from the modified Equation 1:
(2x + 10y - 2z) - (2x - y + 5z + 5z) = -20 - (-9)
2x + 10y - 2z - 2x + y - 10z = -20 + 9
11y - 17z = -11
So, we have a new equation: 11y - 17z = -11. This will be used in the next step.
Step 2: Now let's eliminate the variable x from Equation 3:
Equation 3: 2x - 10y - 5z = 0
To eliminate x, we'll multiply Equation 3 by 1/2:
(1/2)(2x - 10y - 5z) = (1/2)(0)
x - 5y - (5/2)z = 0
Now we can subtract this new equation from Equation 2:
(2x - y + 5z + 5z) - (x - 5y - (5/2)z) = -9 - 0
(2x - x) + (5z + 5z) + (y + 5y) + ((5/2)z - 5y) = -9
x + 10z + 6y - (5/2)z = -9
Simplifying further:
x + (20/2)z + 6y - (5/2)z = -9
x + (10/2)z + 6y = -9
x + 5z + 6y = -9
So, we have a new equation: x + 5z + 6y = -9. This will be used in the final step.
Step 3: Now we have a simplified system of equations:
Equation 4: 11y - 17z = -11
Equation 5: x + 5z + 6y = -9
We can solve this system of equations using different techniques like substitution or graphing. However, if you prefer the elimination method, we can eliminate y from the equations:
Eq 4 × 6: 66y - 102z = -66
Eq 5 × 11: 11x + 55z + 66y = -99
Now subtract the modified Eq 5 from Eq 4:
(66y - 102z) - (11x + 55z + 66y) = -66 - (-99)
66y - 11x - 102z - 55z - 66y = -66 + 99
-11x - 157z = 33
Now we have a new equation: -11x - 157z = 33.
At this point, we have a system of two equations with two variables:
Equation 6: 66y - 102z = -66
Equation 7: -11x - 157z = 33
You can continue solving this system using the elimination method or using other techniques like substitution or matrix methods.
To solve a system of equations by elimination, we want to eliminate one variable by adding or subtracting the equations to create a new equation with one less variable. We will repeat this process until we have a system of equations with only one variable, and then solve for that variable.
Given the system of equations:
1) × + 5y - z = -10
2) 2× - y + 5z + 5z = -9
3) 2× - 10y - 5z = 0
Let's start by eliminating the × variable:
Step 1: Multiply equation 1 by 2 and subtract equation 3 from it to eliminate the × variable.
2(× + 5y - z) - (2× - 10y - 5z) = -10 - 0
2× + 10y - 2z - 2× + 10y + 5z = -10
0 + 20y + 3z = -10
Simplified equation: 20y + 3z = -10 ----(4)
Now, let's eliminate the y variable:
Step 2: Multiply equation 1 by 2 and subtract equation 2 from it to eliminate the y variable.
2(× + 5y - z) - (2× - y + 5z + 5z) = -10 - (-9)
2× + 10y - 2z - 2× + y - 5z - 5z = -10 + 9
11y - 12z = -1
Simplified equation: 11y - 12z = -1 ----(5)
Now, we have two new equations:
4) 20y + 3z = -10
5) 11y - 12z = -1
We can now solve this new system of equations. To do that, we will use a method called substitution or elimination.
Using elimination method:
Step 3: Multiply equation 5 by 20 and equation 4 by 11 to make the coefficients of y in both equations the same.
220y - 240z = -20
220y + 33z = -110
Step 4: Subtract equation 5 from equation 4 to eliminate the y variable.
(220y + 33z) - (220y - 240z) = -110 - (-20)
220y + 33z - 220y + 240z = -110 + 20
273z = -90
Step 5: Solve for z by dividing both sides of the equation by 273.
z = -90 / 273
z ≈ -0.33
Now that we have the value of z, we can substitute it back into equation 4 or 5 to solve for y.
Step 6: Let's substitute z = -0.33 into equation 4.
20y + 3(-0.33) = -10
20y - 0.99 = -10
20y = -10 + 0.99
20y = -9.01
y = -9.01 / 20
y ≈ -0.45
Finally, substitute the values of y and z back into either equation 1, 2, or 3 to solve for ×.
Let's substitute y = -0.45 and z = -0.33 into equation 1:
× + 5(-0.45) - (-0.33) = -10
× - 2.25 + 0.33 = -10
× - 1.92 = -10
× = -10 + 1.92
× ≈ -8.08
Therefore, the solution to the system of equations is:
× ≈ -8.08, y ≈ -0.45, z ≈ -0.33.