Question

solve by substitution method 3x-y=3 and 8x-y=8
plz help me it's urgent..​

Answer 1

$\texttt{\textsf{\large{\underline{Solution}:}}}$

Given equations:

→ 3x - y = 3 – (i)

→ 8x - y = 8 – (ii)

We have to solve the equation by substitution method.

From (i), y = 3x - 3.

Substituting the value of y in (ii), we get:

→ 8x - (3x - 3) = 8

→ 8x - 3x + 3 = 8

→ 5x + 3 = 8

→ 5x = 8 - 3

→ 5x = 5

→ x = 1

Now:

→ y = 3x - 3

→ y = 3 × 1 - 3

→ y = 3 - 3

→ y = 0

Therefore:

$\tt \implies Answer = \begin{cases} \tt x = 1 \\ \tt y = 0 \end{cases}$

$\texttt{\textsf{\large{\underline{Steps To Solve}:}}}$

1. Solve one of the given equations for one of the variables.
2. Substitute the value of the variable in other equation.
3. Solve the resulting single variable equation. Substitute the value of the variable in any of the equations to find the value of the second variable.

Answer 2

$\tt \red{Solution: - }$

$\: \: \: \: \: \: \: \: \: \: \: \: \:$

$\footnotesize \tt{3x - y= 3 - - - - - equation \: 1}$

$\footnotesize \tt{8x - y = 8- - - - - equation \: 2}$

$\: \: \: \: \: \: \: \: \: \: \: \: \:$

$\footnotesize \tt \red{In \: equation \: 1}$

$\: \: \: \: \: \: \: \: \: \: \: \: \:$

$\footnotesize \tt{3x - y= 3 }$

$\footnotesize \tt{x \: = \: } \tt{ \frac{3 + y}{3} }$

$\: \: \: \: \: \: \: \: \: \: \: \: \:$

$\footnotesize \tt \red{In \: equation \: 2}$

$\footnotesize \tt{8x - y = 8}$

$\: \: \: \: \: \: \: \: \: \: \: \: \:$

$\footnotesize \tt \red{Now \: substitute \: the \: value \: of \: x \: in \: eq \: 2}$

$\: \: \: \: \: \: \: \: \: \: \: \: \:$

$\footnotesize\tt{8 } \tt{ \: (\frac{3 + y}{3}) } \footnotesize \tt{ \: - y = 8}$

$\tt{\frac{24 \: + \: 8y}{3}} \footnotesize \tt{ \: - \: y = 8}$

$\tt{\frac{24 \: + \: 8y}{3} - \frac{3y}{3} }\footnotesize \tt{= 8}$

$\tt{\frac{24 \: + \: 8y \: - \: 3y}{3} } \footnotesize \tt{ \: = 8}$

$\tt{\frac{24 \: - \: 5y}{3} }\footnotesize \tt{ \: = 8}$

$\footnotesize \tt{24 \: - \: 5y= 8 \times 3}$

$\footnotesize \tt{24 \: - \: 5y=24}$

$\footnotesize \tt{ - \: 5y=24 - 24}$

$\footnotesize \tt{ - \: 5y=0}$

$\footnotesize \tt{ y \: = \: } \tt{\frac{0}{ - 5} }$

$\boxed{ \underline{ \underline{\footnotesize \tt \red{ y= 0 }}}}$

$\: \: \: \: \: \: \: \: \: \: \: \: \:$

$\footnotesize \tt \red{In \: equation \: 2}$

$\footnotesize \tt{8x - y = 8}$

$\: \: \: \: \: \: \: \: \: \: \: \: \:$

$\footnotesize \tt \red{Now \: substitute \: the \: value \: of \: y\: in \: eq\: 2}$

$\: \: \: \: \: \: \: \: \: \: \: \: \:$

$\footnotesize \tt{8x - 0= 8}$

$\footnotesize \tt{8x = 8 + 0}$

$\footnotesize \tt{8x = 8 }$

$\footnotesize \tt{x } \tt{ \: = \frac{8}{8} }$

$\footnotesize \tt{x } \tt{ \: = \cancel\frac{8}{8} }$

$\boxed{ \underline{ \underline{\footnotesize \tt \red{ x= 1 }}}}$