Question

Solve the pair of equations by the method of substitution:
x+y=a+b ; ax-by=a²-b²​

Answer 1

$\bf{\Huge{\boxed{\rm{\green{ANSWER\::}}}}}$

$\bf{Given}\begin{cases}\tt{x+y=a+b}\\ \tt{ax-by=a^{2} -b^{2} }\end{cases}}$

$\bf{\Large{\underline{\sf{To\:find\::}}}}}$

The value of x & y.

$\bf{\Large{\underline{\rm{\pink{Explanation\::}}}}}$

We have,

• x + y = a + b..........................(1)
• ax - by = a² - b²....................(2)

From equation (1),we get;

$\longmapsto\tt{x+y=a+b}$

$\longmapsto\tt{y=a+b-x............................(3)}$

Putting the value of y in equation (2), we get;

$\longmapsto\tt{ax-b(a+b-x)=a^{2} -b^{2} }$

$\longmapsto\tt{ax-ab-b^{2} +bx=a^{2} -b^{2} }$

$\longmapsto\tt{ax-ab\cancel{-b^{2}} +bx=a^{2} \cancel{-b^{2}} }$

$\longmapsto\tt{ax-ab+bx=a^{2} }$

$\longmapsto\tt{ax+bx=a^{2} +ab}$

$\longmapsto\tt{x(a+b)=a^{2} +ab}$

$\longmapsto\tt{\red{x\:=\:\frac{a^{2}+ab }{a+b} }}$

Putting the value of x in equation (3), we get;

$\longmapsto\tt{y=a+b-(\frac{a^{2} +ab}{a+b}) }$

$\longmapsto\tt{y=a+b-\frac{a^{2} -ab}{a+b} }$

$\longmapsto\tt{y=\frac{(a+b)^{2} -a^{2}-ab }{a+b} }$

$\longmapsto\tt{y=\frac{a^{2}+b^{2} +2ab-a^{2} -ab}{a+b} }$

$\longmapsto\tt{y=\frac{\cancel{a^{2}}+b^{2} +2ab\cancel{-a^{2}} -ab}{a+b} }$

$\longmapsto\tt{\red{y=\frac{b^{2}+ab }{a+b} }}$

Thus,

$\bf{\Large{\boxed{\sf{x=\frac{a^{2}+ab }{a+b} \:\:\:\&\:\:\:y=\frac{b^{2} +ab}{a+b} }}}}}}$

Answer 2

Step-by-step explanation:

We have:-

x + y = a + b

ax - by = a² - b²

To find:-

x and y

Solution:-

x + y = a + b ----------- (1)

ax - by = a² - b² ------ (2)

From equation (1), we can say that

---> x+y = a+b

---> y = a+b-x

[After getting the above, put the value of y in equation 2]

---> ax - b (a+b-x) = a² - b²

---> ax-ab-b²+bx = a² - b²

---> ax-ab+bx = a²

---> ax+bx = a² + ab

---> x (a+b) = a² + ab

---> x = a²+ab / a+b

As we know that,

y = a+b-x

So,

---> y = a+b- (a²+ab / a+b)

---> y = (a+b)² - a² - ab / a+b

---> y = b² +ab / a+b

So, y = b²+ab/a+b

And x = a²+ab/a+b.

Hence, proved.......