To calculate: The solution for the system of provided equations: 5a−2b+3c=10 −3a+b−2c=−7 a+4b−4c=−3

Question

To calculate:  The solution for the system of provided equations:   5a−2b+3c=10  −3a+b−2c=−7  a+4b−4c=−3

Melinda Olson · Accepted Answer

(A) 5a−2b+3c=10  (B) −3a+b−2c=−7  (C) a+4b−4c −3  Consider (A) or (B):  (A) 5a−2b+3c=10  (B) −3a+b−2c=−7  Multiply (B) with 2:  (A) 5a−2b+3c=10  2 ⋅ (B) - 6a + 2b - 4c = -14  Now, add (A) and (B):  (D) −a−c=−4  Consider (A) and (C):  (A) 5a−2b+3c=10  (C) a+4b−4c=−3  Multiply (A) with 2:  2 ⋅ (A) 10a - 4b + 6c = 20  (C) a+4b−4c=−3  Add both the above  equastion:  (E) 11a+2c=17  Now consider (D) and (E):  (D) −a−c=−4  (E) 11a+2c=17  Multipy (D) with 2:  2 ⋅ (D) - 2a - 2c = -8  (E) 11a+2c=17  Now, add both the equastions and solve for a:  9a=9  (dividing both sides by 9)  a=1  Substitute 1 for a in (D):  −(1)−c=−4  Adding 1 to both sides and then dividing by -1,  −c=−3  c=3  Substitude 3 for c and 1 for a in (A) and solve for b:  5(1)−2b+3(3)=10  −2b+14=10  Substracting 14  from both sides and then dividing by -2,  −2b=−4  b=2  −2b=−4  b=2  So, the values obtained are a=1, b=2 and c=3. Substitute thease values  in each of the provided equations to verify:  First Equation: 5(1)−2(2)+3(3)  10  5−4+9  10  10  10  The result is true.  Second Equation:  -3 (1) + (2) - 2(3)  -7  -3 + 2 -6  -7  -7  -7  The result is true.  Third Equation:  (1) + 4 (2) - 4(3)  -3  1 + 8 - 12  -3  -3  -3  The result is true.  5a - 2b + 3c = 10  Therefore, the solution of the system of equations −3a+b−2c=−7 is  a+4b−4c=−3

To calculate: The solution for the system of provided equations:

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