Question

bolve Following equations by using Eliminati
-on mothod
1
x+y=45 : 2x-y= 54
3x+y = 5 ; 2x+3y = 1.​

Answer 1

Answer:

Method used:-

Elimination method

Solution:-

$\mathtt{x + y = 45 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: eq(1)}$

$\mathtt{2x - y = 54 \: \: \: \: \: \: \: \: \: \: \: \: \: eq(2)}$

Now add both equations 1 and 2

$\mathtt{(x + 2x) + (y - y) = 45 + 54}$

$\mathtt{3x = 99}$

$\mathtt{x = \frac{99}{3} }$

$\mathtt{x = 33 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: eq(3)}$

Now substitute the value of x in eq (1)

$\mathtt{x + y = 45}$

$\mathtt{33 + y = 45}$

$\mathtt{y = 45 - 33}$

$\mathtt{y = 12}$

$\fbox\blue{x = 33}$

$\fbox\blue{y = 12}$

Now solve part 2

$\mathtt\blue{3x + y = 5}$

Multiply the whole equation by 2

$\mathtt\blue{2(3x + y = 5)}$

$\mathtt\blue{6x + 2y = 10 \: \: \: \: \: \: \: \: \: \: \: \: \: eq(1)}$

In eq (2) multiply the whole eq by 3

$\mathtt\blue{2x + 3y = 1}$

$\mathtt\blue{3(2x + 3y = 1)}$

$\mathtt\blue{6x + 9y = 3 \: \: \: \: \: \: \: \: \: \: \: \: \: \: eq(2)}$

Subtract the both equation 1 and 2

$\mathtt\blue{(6x - 6x) + (2y - 9y) = 10 - 3}$

$\mathtt\blue{ - 7y = 7}$

$\mathtt\blue{y = - 1 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: eq(3)}$

Now substitute the value of y in equation 1

$\mathtt\blue{6x + 2y = 10}$

$\mathtt\blue{6x + 2 \times ( - 1) = 10}$

$\mathtt\blue{6x - 2 = 10}$

$\mathtt\blue{6x = 10 + 2}$

$\mathtt\blue{6x = 12}$

$\mathtt\blue{x = \frac{12}{6} }$

$\mathtt\blue{x = 2}$

$\fbox{y= \: -1}$

$\fbox{x = 2}$

Step-by-step explanation:

I hope this helps you mate.