Question

3x+y=7 ; 2x-y=5
substitution method
Solve correctly please​

Answer 1

Answer:

Given :-

➲ 3x + y = 7

➲ 2x - y = 5

To Find :-

❒ What is the value of x and y.

Method Used :-

❒ Substitution Method

Solution :-

Given Equation :

$\leadsto \sf\bold{\purple{3x + y =\: 7\: ------\: (Equation\: No\: 1)}}$

$\leadsto \sf\bold{\purple{2x - y =\: 5\: ------\: (Equation\: No\: 2)}}$

From the equation no 2 we get,

$\implies \sf 3x + y =\: 7$

$\implies \sf 3x =\: 7 - y$

$\implies \sf\bold{\purple{x =\: \dfrac{7 - y}{3}\: ------\: (Equation\: No\: 3)}}$

Now, by putting the equation no 3 in the equation no 2 we get,

$\implies \sf 2x - y =\: 5$

$\implies \sf 2\bigg(\dfrac{7 - y}{3}\bigg) - y =\: 5$

$\implies \sf \dfrac{14 - 2y}{3} - y =\: 5$

$\implies \sf \dfrac{14 - 2y - 3y}{3} =\: 5$

By doing cross multiplication we get,

$\implies \sf 14 - 2y - 3y =\: 3 \times 5$

$\implies \sf 14 - 5y =\: 15$

$\implies \sf - 5y =\: 15 - 14$

$\implies \sf - 5y =\: 1$

$\implies \sf\bold{\red{y =\: - \dfrac{1}{5}}}$

Again, by putting the value of y in the equation no 1 we get,

$\implies \sf 3x + y =\: 7$

$\implies \sf 3x + \bigg(- \dfrac{1}{5}\bigg) =\: 7$

$\implies \sf 3x - \dfrac{1}{5} =\: 7$

$\implies \sf 3x =\: 7 + \dfrac{1}{5}$

$\implies \sf 3x =\: \dfrac{35 + 1}{5}$

$\implies \sf 3x =\: \dfrac{36}{5}$

By doing cross multiplication we get,

$\implies \sf 15x =\: 36$

$\implies \sf x =\: \dfrac{\cancel{36}}{\cancel{15}}$

$\implies \sf\bold{\red{x =\: \dfrac{12}{5}}}$

${\small{\bold{\underline{\therefore\: The\: value\: of\: x\: is\: \dfrac{12}{5}\: and\: the\: value\: of\: y\: is\: - \dfrac{1}{5}\: .}}}}$

Answer 2

$\pink{ \boxed{\boxed{\begin{array}{cc} \maltese  \bf \: given \\ \\ \rm \: 3x + y = 7 \: \: \: \: - - - - (1) \\ \\ \rm \: 2x - y = 5 \: \: \: \: - - - - (2) \\ \\ \blue{ {{\begin{array}{cc} \maltese  \bf \: we \: have \: to \: find \: \\ \rm \to \: \: the \: value \: of \: x \: \: and \: \: y \\ \rm by \: substitution \: method \end{array}}}} \end{array}}}}$

Now,

${ \boxed{\boxed{\begin{array}{cc} \maltese  \bf \: from \: eqn.(1) \rm \implies \: \\ \\ \rm \: y = 7 - 3x \: \: \: \: \: \: \: \: - - - - (3) \\ \\ \sf \: put \: the \: value \: of \: y \: in \: eqn.(2)\rm \implies \: \\ \\ \rm \: 2x - (7 - 3x) = 5 \\ \\ \rm \implies \:2x - 7 + 3x = 5 \\ \\ \rm \implies \:5x = 5 + 7 = 12 \\ \\ \rm \implies \:x = \frac{12}{5} \\ \\ \pink{ \boxed{ \rm \therefore \: \: x = \frac{12}{5} }}\end{array}}}}$

${ \boxed{\boxed{\begin{array}{cc} \maltese  \sf \: again \: put \: the \: value \: of \: x \: in \: eqn.(3) \\ \\ \rm \: y = 7 - 3 \times \frac{12}{5} \\ \\ = 7 - \frac{36}{5} \\ \\ = \frac{35 - 36}{5} \\ \\ = \frac{ - 1}{5} \\ \\ \pink{ \boxed{\rm \therefore \: y = - \frac{1}{5} }}\end{array}}}}$

Final answer :

$\orange{ \boxed{\boxed{\begin{array}{cc} \maltese  \bf \: \: value \: of \: \: x = \frac{12}{5} \\ \\ \maltese \bf \: value \: of \: \: y = - \frac{1}{5} \end{array}}}}$