Question

solve the simultaneous equation in substitution Method 2x-3y=4,3y-x=4 ​

Answer 1

Answer:

2x-3y=4    --------1

3y-x=4       --------2

In equation 2

x=3y-4

Substituting x value in 1

2[3y-4]-3y=4

6y-8-3y=4

3y=12

y=4.

so x=12-4

x=8

Therefore the value of x and y is 8 and 4

Answer 2

Answer:

$\Large\bf\purple{Given : } \:$

$\large\mathrm{{2x-3y=4}} \rightarrow \: 1$

$\large\mathrm{ \: 3y - \: x \: = 4} \rightarrow \: 2$

$\Large\bf\purple{To \: \:find : } \: \:$

value of x and y

$\Large\bf\purple{Solution: } \: \: \:$

From equation 2 .

$\large\mathrm{ \: 3y - \: x \: = 4} \: \\ \rightarrow \fbox \mathrm{ x = 3y - 4 }</p><p> \:$

Now Substitute " x = 3y - 4 " in equation 1.

i.e,

$\large\mathrm{{2x-3y=4}}$

$\large\rightarrow \mathrm{ 2 (3y - 4)-3y = 4} \:$

$\large\rightarrow \mathrm{ 6y-8-3y = 4} \\ \\ \large\rightarrow \mathrm{3y - 8 = 4 } \\ \\ \large\rightarrow \mathrm{ 3y = 4+8 } \\ \\ \large\rightarrow \mathrm{ 3y = 12 } \\\\ \large\rightarrow \mathrm{y = \frac{12}{3}} \\ \\ \large\rightarrow \fbox\mathrm{y= 4} \:$

From equation 1.

$\large\mathrm{{2x-3y=4}} \:$

$\large\rightarrow\mathrm{2x-3(4)=4}</p><p>\\ \\ \large\rightarrow \mathrm{2x - 12 = 4} \\ \\ \large\rightarrow \mathrm{2x = 4 +12}\\ \\ \large\rightarrow \mathrm{2x = 16}\\ \\ \large\rightarrow \mathrm{x = \frac{16}{2}}</p><p>$

$\large\rightarrow\fbox{\mathrm{x = 8}} \:$

Hence, the value of x is 8 and y is 4.

$\huge\mathfrak\red{Solved} \:$