Question

solve 2x+3y=11
3x–y=2​

Answer 1

• x = 17/11

• y = 29/11

Step-by-step explanation:

$\huge{\boxed{\bf{Method\ 1}}}$

CROSS MULTIPLICATION METHOD

2x + 3y - 11 = 0  ----(i)

3x - y - 2 = 0​   ----(ii)

Here,

a₁ = 2, b₁ = 3, and c₁ = (-11)

a₂ = 3, b₁ = (-1), and c₁ = (-2)

Now,

$\boxed{\bf{x=\frac{b_{1}c_{2}-b_{2}c_{1}}{a_{1}b_{2}-a_{2}b_{1}}}}$

$\bf{x=\dfrac{[3*(-2)] - [(-1)*(-11)]}{[(2)*(-1)]-[(3)*(3)]}}$

$\bf{x=\dfrac{[-6 - (11)]}{[(-2)-(9)]}}$

$\bf{x=\dfrac{-17}{-11}}$

$\boxed{\bf{x=\frac{17}{11}}}$

Now,

$\boxed{\bf{y=\frac{c_{1}a_{2}-c_{2}a_{1}}{a_{1}b_{2}-a_{2}b_{1}}}}$

$\bf{y=\dfrac{[(-11)*(3)]-[(-2)*(2)]}{[(2)*(-1)]-[(3)*(3)]}}$

$\bf{y=\dfrac{-33-(-4)}{(-2)-(9)}}$

$\bf{y=\dfrac{-33+4}{-11}}$

$\bf{y=\dfrac{-29}{-11}}$

$\boxed{\bf{y=\frac{29}{11}}}$

---------------

$\huge{\boxed{\bf{Method\ 1}}}$

ELIMINATION METHOD

2x + 3y - 11 = 0  ----(i)

3x - y - 2 = 0​   ----(ii)

Multiplying 2 and 3 in eq.(i) and (ii) respectively,

3*(2x + 3y - 11) = 3*0

2*(3x - y - 2​) = 3*0

So,

6x + 9y - 33 = 0   ------(iii)

6x - 2y - 4 = 0  -------(iv)

Now,

By subtraction, we get,

  6x + 9y - 33 = 0  

  6x  - 2y  -  4 = 0

(-)     (+)    (+)

------------------------------

         11y - 29 = 0

------------------------------

Thus,

⇒ 11y -29 = 0

⇒ 11y = 29

⇒ y = 29/11

Now,

• Putting the value of y in any equation, so we take equation (ii)

⇒ 3x - (29/11) - 2 = 0

⇒ (33x - 29 - 22)/11 = 0

[∵ Taking LCM 11 ]

⇒ 33x - 29 - 22 = 11 * 0

⇒ 33x - 29 - 22 = 0

⇒ 33x - 51 = 0

⇒ 33x = 51

⇒ x = 51/33

⇒ x = 17/11

Answer 2

$\bigstar \large {\bf { \pink{Given : - }}}$

• $\bf2x + 3y = 11$

• $\bf \: 3x - y = 2$

$\bigstar \large\bf {\blue{To \: find : - }}$

• $\bf \sf the \: values \: of \: x \: and \: y$

$\bigstar \large{\bf {\green{Solution : - }}}$

$\tt \:2x + 3y = 11 \\ \\ : \implies \tt \: 3y = 11 - 2x \\ \\ : \implies \tt \: y = \frac{11 - 2x}{3} \longrightarrow \: eq(1)$

$\tt3x - y = 2 \\ \\ : \implies \tt \: - y = 2 - 3x \\ \\ : \implies \tt \: - (y) = - (3x - 2) \\ \\ : \tt \implies \: y = 3x - 2 \longrightarrow \: eq(2)$

In Both the equations , eq(1) and eq(2) LHS is equal so RHS Can be equated ,

$: \implies \tt \: \dfrac{11 - 2x}{3} = 3x - 2 \\ \\ \tt : \implies \: 11 - 2x = 3(3x - 2) \\ \\ \tt : \implies \: 11 - 2x = 9x - 6 \\ \\ \tt : \implies \: 11 + 6 = 9x + 2x \\ \\ : \implies \tt \: 17 = 11x \\ \\ \tt : \implies {\underline{\boxed {\tt {x = \frac{17}{11} }}}}$

From eq(2) ,

$\tt : \implies \: y = 3x - 2 \\ \\ : \implies \tt \: y = 3 \bigg( \frac{17}{11} \bigg) - 2 \\ \\ \tt : \implies \: y = \frac{51}{11} - 2 \\ \\ : \implies \tt y = \frac{51 - 22}{11} \\ \\ \tt : \implies {\underline {\boxed {\tt{y = \frac{29}{11} }}}}$

∴ The Values of x and y are $\large\sf{\frac{17}{11} \:and\:\frac{29}{11}}$