Question

plz solve the following equation by substitution method​

Answer 1

Answer:

q=1/3, p = 3/2

Step-by-step explanation:

4p = 3q +5     (1)

p-q = 7/6        (2)

from (2),

p= q+7/6

Substitute value of P in (1)

4(q +7/6) = 3q +5

4q +28/6 = 3q + 5

separating constants and variables we get,

4q-3q = 5-28/6

q = (30-28)/6 = 2/6 or 1/3

put the value of q in (2), we get,

p-1/3 = 7/6

p=7/6 + 1/3

p= 9/6 or 3/2

Answer 2

Answer:

Here, we have given two algebraic equations :

$\dashrightarrow\:\:\sf 4p = 3q + 5$ ⠀⠀..(1)

$\dashrightarrow\:\:\sf p - q = \dfrac{7}{6}$⠀⠀..(2)

So, we are going to use :

${\boxed{\boxed{\sf{\blue{Method \: of \: Elimination \: by \: substitution}}}}}$

In this method, we express one of the variables in terms of the other variable from either of the two equations and then this expression is put in the other equation to obtain an equation in one variable.

$\bigstar \: \large{\sf{\red{Algorithm ( Steps )}}}$

$\bullet \: \:$ Step - 1 Obtain the two equations.

Consider the equations be :

$\implies \sf a_1x + b_1y + c_1 = 0$ --- (1)

and,

$\implies \sf a_2x + b_2y + c_2 = 0$ --- (2)

$\bullet \: \:$ Step - 2 Choose either of the two equations, let us we take equation (1) and find the value of one variable, say y , in terms of the other i.e, x.

$\bullet \: \:$ Step - 3 Substitute the value of y, obtained in Step 2 , in the other equation i.e, (2) to get an equation in x.

$\bullet \: \:$ Step - 4 Solve the equation obtained in step 3 to get the value of x.

$\bullet \: \:$ Step 5 Substitute the value of x obtained in step 4 in the expression for y in terms of x obtained in step 2 to get the value of y.

$\bullet \: \:$ Step 6 The values of x and y obtained in steps 4 and 5 respectively.

____________________________

Taking equation (2)

$\dashrightarrow\:\:\sf p - q = \dfrac{7}{6}$

Transposing p to the R.H.S side , we get :

$\dashrightarrow\:\:\sf - q = - p + \dfrac{7}{6}$

Taking minus common in R.H.S side :

$\dashrightarrow\:\:\sf - q = - (p - \dfrac{7}{6})$

Now, Cancelling negative sign ( - )

$\dashrightarrow\:\:\sf {\not {-}} q = {\not {-}} p - \dfrac{7}{6}$

Now, Plugging (Substituting) the value of ' q ' in the equation (1) in order to get the whole equation in terms of single variable.

$\dashrightarrow\:\:\sf 4p = 3 \times (p - \dfrac{7}{6}) + 5$

$\dashrightarrow\:\:\sf 4p = 3p - \dfrac{21}{6} + 5$

$\dashrightarrow\:\:\sf 4p - 3p = - \dfrac{21}{6} + 5$

$\dashrightarrow\:\:\sf p = \dfrac{- 21}{6} + 5$

$\dashrightarrow\:\:\sf p = \dfrac{- 21 + 30}{6}$

$\dashrightarrow\:\:\sf p = \dfrac{9}{6}$

$\dashrightarrow\:\:\sf p = \dfrac{\not{9}^{ \: \: 3} }{\not{6}^{ \: \: 2} }$

$\dashrightarrow\:\:\bf p = \dfrac{3}{2}$

$\dashrightarrow\:\: \underline{ \boxed{\sf The \: Value \: of \: p = \dfrac{3}{2} }}$

Now, Substituting the value of ' p ' in equation (2)

$::\implies \:\sf p - q = \dfrac{7}{6}$

$::\implies \:\sf \dfrac{3}{2} - q = \dfrac{7}{6}$

$::\implies \:\sf \dfrac{3}{2} - \dfrac{7}{6} = q$

Taking L.C.M :-

$::\implies \:\sf \dfrac{9 - 7}{6} = q$

$::\implies \:\sf \dfrac{2}{6} = q$

$::\implies \:\sf q = \dfrac{\not{2}^{ \: \: 1} }{\not{6}^{ \: \: 3} }$

$::\implies \:\bf q = \dfrac{1}{3}$

Hence,

$\dashrightarrow\:\: \underline{ \boxed{\sf The \: Value \: of \: q = \dfrac{1}{3} }}$

_______________________

Checking the answer :-

Verification :

Putting $\bf p = \dfrac{3}{2} \: and \: q = \dfrac{1}{3}$ respectively in Equation (2)

$\dashrightarrow\:\:\sf \dfrac{3}{2} - \dfrac{1}{3} = \dfrac{7}{6}$

Taking L.C.M :

$\dashrightarrow\:\:\sf \dfrac{9 - 2}{6} = \dfrac{7}{6}$

$\dashrightarrow\:\:\sf \dfrac{7}{6} = \dfrac{7}{6}$

$\dashrightarrow\:\:\sf L.H.S = R.H.S$

Hence,

${\boxed {\sf{Verified ! }}}$

_________________________

What we have done ?

• Here , we use the method of Elimination by substitution. We take second equation and consider the value of one variable. Thereafter , we substitute the value in the equation (2) in order to get the value of second variable. Then, we again substitute the value of second variable in first. That's how we got the values of p and q respectively.