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5 a 6 b 3 c = 4 4 a 2 b 4 c = 6 2 a + 6 b 3 c = 22 {:[5a-6b-3c=4],[4a-2b-4c=-6],[-2a+6b-3c=-22]:}
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To solve this system of equations, we can use the method of elimination or substitution. Here, I will show you the elimination method.
First, let’s eliminate the variable a a. We can do this by multiplying the second equation by 5 5 and the third equation by 1 -1, and then adding them to the first equation. This will eliminate a a:
5 ( 4 a 2 b 4 c ) + ( 1 ) ( 2 a + 6 b 3 c ) = 5 ( 6 ) + ( 1 ) ( 22 ) 5(4a-2b-4c)+(-1)(-2a+6b-3c)=5(-6)+(-1)(-22)
Simplifying, we get:
20 a 10 b 20 c + 2 a 6 b + 3 c = 30 + 22 20 a-10 b-20 c+2a-6b+3c=-30+22
Combining like terms, we have:
22 a 16 b 17 c = 8 22 a-16 b-17 c=-8
Next, let’s eliminate the variable b b. We can do this by multiplying the first equation by 2 2 and the third equation by 3 3, then adding them to the second equation:
2 ( 5 a 6 b 3 c ) + 3 ( 2 a + 6 b 3 c ) = 2 ( 4 ) + 3 ( 2 ) 2(5a-6b-3c)+3(-2a+6b-3c)=2(4)+3(-2)
Simplifying, we get:
10 a 12 b 6 c 6 a + 18 b 9 c = 8 6 10 a-12 b-6c-6a+18 b-9c=8-6
Combining like terms, we have:
4 a + 6 b 15 c = 2 4a+6b-15 c=2
Now we have two equations:
22 a 16 b 17 c = 8 22 a-16 b-17 c=-8
4 a + 6 b 15 c = 2 4a+6b-15 c=2
To eliminate the variable b b, let’s multiply the first equation by 3 3 and the second equation by 8 8, then subtract the equations:
3 ( 22 a 16 b 17 c ) 8 ( 4 a + 6 b 15 c ) = 8 16 3(22 a-16 b-17 c)-8(4a+6b-15 c)=-8-16
Simplifying, we get:
66 a 48 b 51 c 32 a 48 b + 120 c = 24 16 66 a-48 b-51 c-32 a-48 b+120 c=-24-16
Combining like terms, we have:
34 a 96 b + 69 c = 40 34 a-96 b+69 c=-40
Now we have two equations:
34 a 96 b + 69 c = 40 34 a-96 b+69 c=-40
4 a + 6 b 15 c = 2 4a+6b-15 c=2
To eliminate the variable a a, let’s multiply the second equation by 17 17 and the first equation by 4 4, then subtract the equations:
17 ( 4 a + 6 b 15 c ) 4 ( 34 a 96 b + 69 c ) = 2 ( 40 ) 17(4a+6b-15 c)-4(34 a-96 b+69 c)=2-(-40)
Simplifying, we get:
68 a + 102 b 255 c 136 a + 384 b 276 c = 42 + 80 68 a+102 b-255 c-136 a+384 b-276 c=42+80
Combining like terms, we have:
68 a + 486 b 531 c = 122 -68 a+486 b-531 c=122
Now we have two equations:
34 a 96 b + 69 c = 40 34 a-96 b+69 c=-40
68 a + 486 b 531 c = 122 -68 a+486 b-531 c=122
We can solve for a a in the first equation:
a = 96 b + 69 c 40 34 a=(-96 b+69 c-40)/(34)
Substituting this into the second equation, we get:
68 ( 96 b + 69 c 40 34 ) + 486 b 531 c = 122 -68((-96 b+69 c-40)/(34))+486 b-531 c=122
Simplifying this expression, we get:
96 b 69 c 80 + 486 b 531 c = 122 96 b-69 c-80+486 b-531 c=122
Combining like terms, we have:
582 b 600 c = 202 582 b-600 c=202
To solve for b b, we can divide this equation by 582 582:
b = 600 c + 202 582 b=(600 c+202)/(582)
Finally, we can substitute the values of a a and b b back into any of the original equations to solve for c c. Let’s use the first equation:
5 a 6 b 3 c = 4 5a-6b-3c=4
After simplifying and solving for c c, we find:
c = 415 582 c=(415)/(582)
Substituting this value back into the equation for b b, we get:
b = 203 291 b=(203)/(291)
And substituting the values of b b and c c back into the equation for a a, we get:
a = 115 291 a=(-115)/(291)
Therefore, the solution to the system of equations is:
a = 115 291 a=(-115)/(291), b = 203 291 b=(203)/(291), c = 415 582 c=(415)/(582)
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