A solution x- y of the system of equations x-y- 13 and x2-y2-4 is given by

Consider the given equations. x y=13 #160; #160; #160; #160; #160; #160; ............(1) x^2 y^2=4 (x y)(x+y)=4⇒ x+y=12 ...

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Standard X

Mathematics

Cross-Multiplication Method of Finding Solution of a Pair of Linear Equations

A solution ...

Question

A solution $(x, y)$ of the system of equations $x - y = \frac{1}{3}$ and $x^{2} - y^{2} = 4$ is given by

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Solution

Consider the given equations.
$x - y = \frac{1}{3}$ $. . . . . . . . . . . . (1)$

$x^{2} - y^{2} = 4$

$(x - y) (x + y) = 4 \Rightarrow x + y = 12$ $. . . . . . . . . . . . (2)$

On solving equations $(1)$ and $(2)$ , we get

$x = \frac{37}{6}$ and $y = \frac{35}{6}$

Hence, this is the answer.

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A solution (x,y) of the system of equations x−y=13 and x2−y2=4 is given by

Consider the given equations.x−y=13 ............(1)x2−y2=4(x−y)(x+y)=4⇒x+y=12 ............(2)On solving equations (1) and (2), we getx=376 and y=356Hence, this is the answer.

A solution $(x, y)$ of the system of equations $x - y = \frac{1}{3}$ and $x^{2} - y^{2} = 4$ is given by

Consider the given equations.
$x - y = \frac{1}{3}$ $. . . . . . . . . . . . (1)$

$x^{2} - y^{2} = 4$

$(x - y) (x + y) = 4 \Rightarrow x + y = 12$ $. . . . . . . . . . . . (2)$

On solving equations $(1)$ and $(2)$ , we get

$x = \frac{37}{6}$ and $y = \frac{35}{6}$

Hence, this is the answer.