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Worked Examples · Example 40

Q.Solve the following system of equation using Cramer's rule: 2x−3y=52x - 3y = 5; 5x+3y=25x + 3y = 2.

Andaman Nicobar CbseNCERTSubjective· 3mImportance★★★★★
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✓ Free question

Cramer's rule with D=21, Dx=21, Dy=−21D=21,\ D_x=21,\ D_y=-21 gives x=1, y=−1x=1,\ y=-1.

For a1x+b1y=c1, a2x+b2y=c2a_1x+b_1y=c_1,\ a_2x+b_2y=c_2: x=DxD, y=DyDx=\dfrac{D_x}{D},\ y=\dfrac{D_y}{D}, where D=∣a1b1a2b2∣D=\begin{vmatrix}a_1&b_1\\a_2&b_2\end{vmatrix}, and Dx,DyD_x,D_y replace the respective column with the constants.

  1. Equations: 2x−3y=5, 5x+3y=22x-3y=5,\ 5x+3y=2.

  2. Main determinant:

D=∣2−353∣=(2)(3)−(−3)(5)=6+15=21.D = \begin{vmatrix} 2 & -3 \\ 5 & 3 \end{vmatrix} = (2)(3)-(-3)(5) = 6+15 = 21.

  1. DxD_x (replace column 1 with constants):

Dx=∣5−323∣=(5)(3)−(−3)(2)=15+6=21.D_x = \begin{vmatrix} 5 & -3 \\ 2 & 3 \end{vmatrix} = (5)(3)-(-3)(2) = 15+6 = 21.

  1. DyD_y (replace column 2 with constants):

Dy=∣2552∣=(2)(2)−(5)(5)=4−25=−21.D_y = \begin{vmatrix} 2 & 5 \\ 5 & 2 \end{vmatrix} = (2)(2)-(5)(5) = 4-25 = -21.

  1. Therefore

x=DxD=2121=1,y=DyD=−2121=−1.x = \frac{D_x}{D} = \frac{21}{21} = 1, \qquad y = \frac{D_y}{D} = \frac{-21}{21} = -1.

  1. Check: 2(1)−3(−1)=52(1)-3(-1)=5 ✓; 5(1)+3(−1)=25(1)+3(-1)=2 ✓.
✓Final answer

x=1,y=−1x = 1,\quad y = -1

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