
To solve: The given system of linear equations and then to verify the answer.

Answer to Problem 3CT
The solution is (26,−3,15) .
Explanation of Solution
Given information: A system of linear equations.
Calculation:
−2x+y+4z=5 −(i) x+3y−z=2 −(ii) 4x+y−6z=11 −(iii)
Multiplying (ii) by 2 and adding it to (i) ,
2x+6y−2z=4−2x+y+4z =5============== 7y+2z=9 −(iv)
Multiplying (ii) by 4 and subtracting (iii) from it,
4x+12y−4z=84x+y−6z =11============= 11y+2z=−3 −(v)
Subtracting (iv) from (v) ,
11y+2z=−37y+2z =9========== 4y=−12 y=−3
Substituting derived value y of in (iv) ,
7(−3)+2z=9 2z=30 z=15
Substituting derived values of z and y in (ii) ,
x+3(−3)−15=2 x=26
Verifying by putting values of x,y and z in (i),(ii) and (iii) ,
−2(26)−3+4(15)=5 5+60=5 5=5 L.H.S.=R.H.S.26+3(−3)−15=2 26−9−15=2 2=2 L.H.S.=R.H.S.4(26)−3−6(15)=11 104−3−90=11 11=11 L.H.S.=R.H.S.
Hence, verified.
Thus, the solution is (26,−3,15) .
Chapter 1 Solutions
Big Ideas Math A Bridge To Success Algebra 2: Student Edition 2015
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