Problem 280 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

Solve \(x+3 y-2 z=0\) \(2 x-y+4 z=0\) \(x-11 y+14 z=0\)

Step-by-Step Solution

Answer

After performing the above steps, the ratio of \(x : y : z\) turns out to be \(12 : \frac{8}{7} : 1\). In other words, \(x = 12k, y = \frac{8k}{7}, z = k\) for an arbitrary number \(k\).

Step 1: Equating the first and second equation

Subtract the first equation \(x+3y-2z=0\) from the second equation \(2x-y+4z=0\), This gives \(x - 4y + 6z = 0\)

Step 2: Equating the first and third equation

Subtract the first equation \(x+3y-2z=0\) from the third equation \(x-11y+14z=0\), This gives \( -14y + 16z = 0\)

Step 3: Solving for y

From \(-14y + 16z = 0\), we can derive the value of \(y\) as \(y = \frac{8}{7}z\)

Step 4: Substituting y in Step 1 equation

Substitute value of \(y\) from Step 3 into the equation derived in Step 1 i.e \(x - 4y + 6z = 0\) to get value of \(x\) = \(12z\)

Step 5: Substituting x and y in original equation

Substitute \(x = 12z\) and \(y = \frac{8}{7}z\) in any of the original equations to find the ratio of \(z\)