Question

Find the value of x and y by elimination method ​

Answer 1

Step-by-step explanation:

Given -

x/4 + y = -5

2x/3 + y/2 = -7

To Find -

• Value of x and y

By elimination method :-

→ [ x/4 + y = -5 ] × 1/2

[ 2x/3 + y/2 = -7 ] × 1

→ x/8 + y/2 = -5/2

2x/3 + y/2 = -7

(-) (-) (+)

________________

→ x/8 - 2x/3 = 7 -5/2

→ 3x - 16x/24 = 14-5/2

→ -13x/24 = 9/2

→ x = 9×24/2×-13

→ x = -12×9/13

→ x = -108/13

Now, Substituting the value of x on x/4 + y = -5, we get :

→ -108/13×4 + y = -5

→ -108/52 + y = -5

→ y = -5 + 108/52

→ y = -260 + 108/52

→ y = -152/52

Hence,

The value of x is -108/13 and y is -152/52

Verification :-

• x/4 + y = -5

→ -108/52 - 152/52 = -5

→ -260/52 = -5

→ -5 = -5

LHS = RHS

And

• 2x/3 + y/2 = -7

→ 2×-108/3×13 + -152/52×2 = -7

→ -216/39 - 76/52 = -7

→ -864 - 228/156 = -7

→ -1092/156 = -7

→ -7 = -7

LHS = RHS

Hence,

Verified..

Answer 2

$\large{\boxed{\underline{\underline{\bf{\red{Answer:-}}}}}}$

$\implies$ x = -108/13

$\implies$ y = -152/52

$\large\underline\mathrm{Given:-}$

• x/4 + y = -5
• 2x/3 + y/2 = -7

$\large\underline\mathrm{To \: find}$

• Value of x and y

$\large\underline\mathrm{By \: Elimation \: method:}$

• x/4 + y = -5. ....(1)
• 2x/3 + y/2 = -7. ...(2)

The equation (1). × (1/2) and (2) × (1)

$\implies$ [x/8 + y/2 = -5/2] × (1/2)

$\implies$ [2x/3 + y/2 = -7] × 1

$\implies$ 3x - 16x/24 = 14 - 5/2

$\implies$ -13x/24 = 9/2

$\implies$ x = 9 × 24/2 × -13

$\implies$ x = -12×9/3

$\implies$ x = -108/13

$\large\underline\mathrm{Now,}$

$\large\underline\mathrm{Substitution \: \: the \: value \: of \: x \: on \: x \: / \: 4 \: + \: y \: = \: -5 \: , \: we \: get}$

$\implies$ -108/13x × 4 + y = -5

$\implies$ 108/52 × 4 + y = -5

$\implies$ y = -5 + 108/52

$\implies$ y = (-260 + 108)/52

$\implies$ y = -152/52

$\large\underline\mathrm{hence,}$

$\large\underline\mathrm{The \: value \: of \: x \: is. \: -108 \: / \: 13 \: and \: y \: is \: -152 \: / \: 52}$

$\large\underline\mathrm{Hope \: it \: helps \: you \: plz \: mark \: me \: brainlist}$