Question

Solve the following:
\[\begin{align}
  & ax+by+\left( -c \right)=0 \\
 & bx+ay=1+c \\
\end{align}\]

Answer 1

Hint: In this question, we have been asked to solve for x and y and when we are asked to solve for x and y, we think of a few methods like grouping, substitution, or elimination method. In this question, the grouping method is not possible because we are given coefficients in terms of a, b and c. So, we will use the substitution method

Complete Step-by-step answer:
In this question, we are asked to solve the equations for x and y where the equations are ax + by + (– c) = 0 and bx + ay = 1 + c. To solve this question, we will use the substitution method and for that, we will find the value of one variable in terms of other variables from one equation and will put that in another equation. So, we will consider the equation ax + by + (– c) = 0 first and then we will find the value of x in terms of y. So, we can write,
\[ax+by+\left( -c \right)=0\]
\[ax+by=c\]
\[ax=c-by\]
Now, we will divide the whole equation by a. So, we will get,
\[x=\dfrac{c-by}{a}.....\left( i \right)\]
Now, we will put the value of x from equation (i) in bx + ay = 1 + c. Therefore, we will get,
\[b\left( \dfrac{c-by}{a} \right)+ay=1+c\]
Now, we will simplify it to find the value of y. So, we can write
\[\dfrac{bc-{{b}^{2}}y}{a}+ay=1+c\]
Now, we will take the LCM. So, we will get,
\[\dfrac{bc-{{b}^{2}}y+{{a}^{2}}y}{a}=1+c\]
Now, we will cross multiply the equation. So, we will get,
\[bc+\left( {{a}^{2}}-{{b}^{2}} \right)y=a+ac\]
Now, we will keep the variable y on LHS and rest on RHS
\[\left( {{a}^{2}}-{{b}^{2}} \right)y=a+ac-bc\]
\[y=\dfrac{a+c\left( a-b \right)}{{{a}^{2}}-{{b}^{2}}}....\left( ii \right)\]
Now, from equation (ii), we will put the value of y in equation (i). So, we get,
\[x=\dfrac{c-b\left[ \dfrac{a+c\left( a-b \right)}{{{a}^{2}}-{{b}^{2}}} \right]}{a}\]
Now we will simplify it further to find the value of x. So, we get,
\[x=\dfrac{c-\dfrac{ba+bc\left( a-b \right)}{{{a}^{2}}-{{b}^{2}}}}{a}\]
Now, we will take the LCM of the terms in the numerator. So, we get,
\[x=\dfrac{\dfrac{{{a}^{2}}c-{{b}^{2}}c-\left( ba+abc-{{b}^{2}}c \right)}{{{a}^{2}}-{{b}^{2}}}}{a}\]
\[x=\dfrac{{{a}^{2}}c-{{b}^{2}}c-ab-abc+{{b}^{2}}c}{a\left( {{a}^{2}}-{{b}^{2}} \right)}\]
Now, we know that similar terms get canceled out. So, we get,
\[x=\dfrac{{{a}^{2}}c-ab-abc}{a\left( {{a}^{2}}-{{b}^{2}} \right)}\]
Now, we can see that ‘a’ is common in all the terms of the numerator. So, we get,
\[x=\dfrac{a\left( ac-b-bc \right)}{a\left( {{a}^{2}}-{{b}^{2}} \right)}\]
And we know that common terms get canceled out, so we get,
\[x=\dfrac{c\left( a-b \right)-b}{{{a}^{2}}-{{b}^{2}}}....\left( iii \right)\]
From equation (ii) and (iii), we can say that \[x=\dfrac{c\left( a-b \right)-b}{{{a}^{2}}-{{b}^{2}}}\] and \[y=\dfrac{a+c\left( a-b \right)}{{{a}^{2}}-{{b}^{2}}}\] are the solution of the given pair of the equation.

Note: In this question, we can also use the elimination method to find the value of x and y by somehow forming the equation in one variable, and then after finding the value of that variable, we will use that value to find the value of the other variable.