Question 177760
<FONT FACE="Times New Roman" SIZE="4">I'll do the first one of these and you can repeat the process for the other one.


Equation 1: *[tex \Large \text {  } \math 2a + b = 5]
Equation 2: *[tex \Large \text {  } \math a - 2b = 10]


Step 1:  Multiply Eq. 1 by 2 so that the coefficients on the b terms are additive inverses (one is already -2 and the other becomes 2):


Equation 3: *[tex \Large \text {  } \math 4a + 2b = 10]


Step 2:  Add Eq 3 to Eq 2, term by term, eliminating the b variable (hence the name of the method)


Equation 4: *[tex \Large \text {  } \math 4a + a + 2b - 2b = 10 + 10 \text{ } \Rightarrow \text{ } \math 5a = 20 \text{ } \Rightarrow \text{ } \math a = 4]


Step 3:  Substitute the value determined for a into either of the original equations and then solve for b:


Equation 2: *[tex \Large \text {  } \math 4 - 2b = 10\text{ } \Rightarrow \text{ } \math -2b = 6 \text{ } \Rightarrow \text{ } \math b = -3]


Hence, the solution set for the system:


Equation 1: *[tex \Large \text {  } \math 2a + b = 5]
Equation 2: *[tex \Large \text {  } \math a - 2b = 10]


is the ordered pair *[tex \Large (a, b) = (4, -3)]


Checking the answer is left as an exercise for the student.
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