Question

# In the system of equation 1x+1y+1z=6,2x−3y+4z=8 and 3x−4y+5z=10, values of x, y and z which satisfy the above three equations, will be

A
12, 1 and 13
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B
13, 1 and 12
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C
1,12 and 13
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D
1,13 and 12
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Solution

## The correct option is C 1,12 and 13If 1x=A,1y=B and 1z=C, then the given equation will be written as:A+B+C=6.........(i)2A−3B+4C=8 ...............(ii)3A−4B+5C=10 ..... (iii)On multiplying equation (i) by 3 and adding it to equation (ii),(3A+3B+3C)+(2A−3B+4C)=6×3×85A+7C=26 ...... (iv)On multiplying equation (i) by 4 and adding it to equation (iii),(4A+4B+4C)+(3A−4B+5C)=6×4+107A+9C=34 ..... (v)On multiplying equation (iv) by 7 and equation(v) by 5 and subtracting equation (v) from equation (iv),(35A+49C)−(35A+45C)=26×7−34×54C=182−170=12⇒C=124=3On substituting this value of C in equation (iv),5a+7×3=26⇒5A=26−21∴5A=5⇒A=55=1From equation (i), 1+B+3=6⇒B=6−4=2∵1x=A=1⇒x=1,1y=B=2⇒y=12and 1z=C=3⇒z=13Hence values of x, y and z will be 1,12 and 13 respectively.

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