Question

how to solve this equation by cramer's rule 2x+3y=2;x-y/2=1/2

Answer 1

we know, if two equations a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 are given then,
Solution of this equation with help of Cramer's rule is given by
x/(b₁c₂ - b₂c₁) = -y/(a₁c₂ - a₂c₁) = 1/(a₁b₂ - a₂b₁)

Now, use this concept here,
We have equations :
2x + 3y = 2 ⇒2x + 3y - 2 = 0
x - y/2 = 1/2 ⇒2x - y - 1 = 0

x/{3 × (-1) - (-1) × (-2)} = -y/{2 × (-1) - 2 × (-2)} = 1/{2 × (-1) - 2 × 3}
x/(-3 -2) = -y/(-2 + 4) = 1/(-2 - 6)
x/-5 = -y/2 = 1/-8
x/5 = y/2 = 1/8
∴ x = 5/8 and y = 2/8 = 1/4

Hence answer is x = 5/8 And y = 1/4

Answer 2

The given equations are as follows-
(1) $2x+3y=2$
(2) $x- \frac{y}{2}= \frac{1}{2}$

By Cramer's Rule,

$I = \left|\begin{array}{ccc}2&3\\1& -\frac{1}{2} \end{array}\right| = -4$

$I_{x} = \left|\begin{array}{ccc}2&3\\\frac{1}{2}& -\frac{1}{2} \end{array}\right| = -2.5$

$I_{y} = \left|\begin{array}{ccc}2&2\\1& \frac{1}{2} \end{array}\right| = -1$

As I ≠ 0, the given set of equations are consistent and have an independent solution.

Hence, by Cramer's Rule,
$x= \frac{ I_{x} }{I} = \frac{-2.5}{-4} = 0.625 = \frac{5}{8}$
$y= \frac{ I_{y} }{I} = \frac{-1}{-4} = 0.25 = \frac{1}{4}$

So, the solution set is (0.625, 0.25)

[ANSWERED]