2a+3b-4c=7,
a-b+2c=6
a) find an ordered triple of numbers (a,b,c) that satisfies both equations
b) can you find a second ordered triple that satisfies both equations
c) solve for b in terms of a, and solve for c in terms of a. how many solutions does this system have?
i dont understand how i can solve it because there are 3 variables but only 2 equations?
for c) i got
b=6+2c+a
i dont know how to transform the c in a term? please help
let's do c) first:
first equation as is ---> 2a + 3b - 4c = 7
double the second --> 2a - 2b + 4c = 12
now add them:
4a + b = 19
b = 19 - 4a
now you can pick any value for a that you want,
e.g. a = 2
then b = 19-8 = 11
sub those values back into the original second equation:
a-b+2c=6
2-11+2c = 6
2c = 15
c = 7.5 , so a solution would be (2, 11, 7.5)
how about a=0 , then
b = 19-0 = 19
in a-b+2c=6
0 - 19 + 2c = 6
c = 12.5
etc
There will be an infinite number of solutions.
suppose you are given x + y = 10
you are given 2 variables but only one solution, so there would be an infinite number of solutions, namely all the points on the given line.
Your question has the same concept, except each of the given equations could represent the equation of a plane in 3D.
So you are really looking at the intersection of two planes. Think of the case where a wall meets the ceiling, there would be an infinite number of points on that line of intersection.
thank you very much! i understand now
To solve a system of linear equations, you need to have as many equations as variables. In this case, you have two equations and three variables, so there are infinitely many solutions possible. Let's go through the steps to find the solutions:
a) To find an ordered triple (a, b, c) that satisfies both equations, we can use the method of substitution or elimination. Let's use the method of elimination:
Multiply the first equation by 2 and the second equation by 3 to make the coefficients of 'c' the same:
2(2a + 3b - 4c) = 2(7) -> 4a + 6b - 8c = 14
3(a - b + 2c) = 3(6) -> 3a - 3b + 6c = 18
Now, add the two equations together to eliminate 'c':
(4a + 6b - 8c) + (3a - 3b + 6c) = 14 + 18
7a + 3b - 2c = 32
Next, let's solve the two-equation system:
2a + 3b - 4c = 7
7a + 3b - 2c = 32
To eliminate 'b', multiply the first equation by 3 and the second equation by -1:
3(2a + 3b - 4c) = 3(7) -> 6a + 9b - 12c = 21
-1(7a + 3b - 2c) = -1(32) -> -7a - 3b + 2c = -32
Add the two equations together to eliminate 'b':
(6a + 9b - 12c) + (-7a - 3b + 2c) = 21 + (-32)
-a + 6c = -11
Now we have the two equations:
-a + 6c = -11
7a + 3b - 2c = 32
We can solve the first equation for 'a':
a = 6c - 11
b) To find a second ordered triple that satisfies both equations, we can choose any value for 'c' and then determine the values of 'a' and 'b' using the equations. Let's choose c = 0:
a = 6(0) - 11 = -11
Substituting a = -11 into the second equation:
7(-11) + 3b - 2(0) = 32
-77 + 3b = 32
3b = 109
b = 109/3
So, a second ordered triple that satisfies both equations is (-11, 109/3, 0).
c) To solve for 'b' in terms of 'a' and 'c', we can substitute the value of 'a' from the first equation into the second equation:
a - b + 2c = 6
(6c - 11) - b + 2c = 6
8c - b = 17
So, we can write 'b' in terms of 'a' and 'c' as:
b = 8c - 17
To solve for 'c' in terms of 'a', we can rearrange the first equation:
2a + 3b - 4c = 7
4c = 2a + 3b - 7
c = (2a + 3b - 7)/4
The system has infinitely many solutions because the number of variables is greater than the number of independent equations.
To solve a system of linear equations like this one with three variables but only two equations, we will use a method called substitution.
a) To find an ordered triple of numbers (a,b,c) that satisfies both equations, we will solve the equations simultaneously. Start by solving one equation for one variable and substitute it into the other equation.
From equation 2a + 3b - 4c = 7, solve for a:
2a = 7 - 3b + 4c
a = (7 - 3b + 4c) / 2
Now, substitute this value of a into the second equation: a - b + 2c = 6
((7 - 3b + 4c) / 2) - b + 2c = 6
Next, simplify the equation and solve for either b or c:
7 - 3b + 4c - 2b + 4c = 12
7 - 5b + 8c = 12
-5b + 8c = 5
Now, we have two equations:
-5b + 8c = 5 (from substituting a)
2a + 3b - 4c = 7 (original equation)
Use these two equations to solve for b and c.
You can use either substitution or elimination methods to solve the system of equations.
b) To find a second ordered triple that satisfies both equations, you can come up with a different set of b and c values that still satisfy the equations. You can choose any values for b and c, substitute them into the equations, and then solve for a using the equations again.
c) To solve for b in terms of a, and solve for c in terms of a, we can rearrange the equations.
From equation 2a + 3b - 4c = 7, solve for b:
3b = 7 - 2a + 4c
b = (7 - 2a + 4c) / 3
From equation a - b + 2c = 6, solve for c:
2c = 6 - a + b
c = (6 - a + b) / 2
Now, you have expressions for b in terms of a and c in terms of a. The system has infinitely many solutions because for every value you choose for a, you can find corresponding values for b and c that satisfy the equations.