Solve the following pair of equations: (a neq | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given equations are:$\frac{2}{a} + \frac{3}{b} = -1$  --- $(1)$$\frac{4}{a} - \frac{9}{b} = -7$  --- $(2)$Let $\frac{1}{a} = x$ and $\frac{1}{b} =

Vedclass · Accepted Answer

$(-1, 3)$

Vedclass · Answer

(A) Given equations are:$\frac{2}{a} + \frac{3}{b} = -1$  --- $(1)$$\frac{4}{a} - \frac{9}{b} = -7$  --- $(2)$Let $\frac{1}{a} = x$ and $\frac{1}{b} = y$. Then the equations become:$2x + 3y = -1$  --- $(3)$$4x - 9y = -7$  --- $(4)$Multiply equation $(3)$ by $3$ to eliminate $y$:$6x + 9y = -3$  --- $(5)$Adding equation $(4)$ and $(5)$:$(4x - 9y) + (6x + 9y) = -7 + (-3)$$10x = -10$$x = -1$Since $x = \frac{1}{a}$,we have $\frac{1}{a} = -1$,so $a = -1$.Substitute $x = -1$ into equation $(3)$:$2(-1) + 3y = -1$$-2 + 3y = -1$$3y = 1$$y = \frac{1}{3}$Since $y = \frac{1}{b}$,we have $\frac{1}{b} = \frac{1}{3}$,so $b = 3$.Thus,the solution is $(a, b) = (-1, 3)$.

Solve the following pair of equations: $(a \neq 0, b \neq 0)$
$\frac{2}{a} = -1 - \frac{3}{b}$
$\frac{9}{b} = \frac{4}{a} + 7$

Similar Questions

Solve the following pair of linear equations: $2(3u - v) = 5uv$ and $2(u + 3v) = 5uv$.

Find the value$(s)$ of $p$ for the pair of equations $2x + 3y - 5 = 0$ and $px - 6y - 8 = 0$ if the system has a unique solution.

The solution of the pair of equations $x+2=0$ and $y-1=0$ is $(x, y) = \dots$

The angles of a cyclic quadrilateral $ABCD$ are $\angle A = (6x + 10)^{\circ}$,$\angle B = (5x)^{\circ}$,$\angle C = (x + y)^{\circ}$,and $\angle D = (3y - 10)^{\circ}$. Find $x$ and $y$,and hence the values of the four angles.

The pair of equations $5x - y = 9$ and $10x - 18 = 2y$ is $\ldots \ldots \ldots \ldots$

Vedclass Test Series

Exam Paper Generator

Online Exam Module

Solve the following pair of equations: $(a \neq 0, b \neq 0)$$\frac{2}{a} = -1 - \frac{3}{b}$$\frac{9}{b} = \frac{4}{a} + 7$