Equations Reducible to a Pair of Linear Equations in Two Variables
Example : A boat goes 30 km upstream and 44 km downstream in 10 hours. In 13 hours, it can go
40 km upstream and 55 km down-stream. Determine the speed of the stream and that of the boat in still water.
In this section, we shall discuss the solution of such pairs of equations which are not linear but can be reduced to linear form by making some suitable substitutions.
We now explain this process through some examples.
Example: Solve the pair of equations:
$${2/x+3/y=13}$$
$${5/x-4/y= -2}$$
Solution : Write the given pair of equations as
$${2 (1/x)+3 (1/y)=13}$$
$${5 (1/x)-4 (1/y)= -2}$$
These equations are not in the form ax + by + c = 0.
However, if we substitute ${1/x=p}$ and ${1/y=q}$;
we get ;
$${2p+3q=13}$$
$${5p-4q= -2}$$
So, we have expressed the equations as a pair of linear equations. Now, you can use any method to solve these equations, and get p = 2, q = 3.
Since ${1/x=p}$ and ${1/y=q}$; Substitute the values of p and q
${1/x=2}$ and ${1/y=3}$
Therefore ${x = 1/2}$ and ${y = 1/3}$
Verification : By substituting ${x = 1/2}$ and ${y = 1/3}$ in the given equations, we find that
both the equations are satisfied.
Example : Solve the following pair of equations by reducing them to a pair of linear equations : $${5/(x-1) + 1/(y-2) = 2}$$ $${6/(x-1) - 3/(y-2) = 1}$$Solution : Let ${1/(x-1)=p}$ and ${1/(y-2)=q}$ Subsitute it in the above equations: ${5p+1q=2}$ (3) ${6p-3q=1}$ (4) Now, use any method to solve these equations; we get ${p=1/3}$ and ${q=1/3}$ Now, substituting ${1/(x-1)=p}$ , we get ${1/(x-1)=1/3}$ ; ${x=4}$ Now, substituting ${1/(y-2)=q}$ , we get ${1/(y-2)=1/3}$ ; ${y=5}$ Hence, ${x=4}$, ${y=5}$ is the required solution of the given pair of equations.
Solution : Let the speed of the boat in still water be ${x}$ km/h and speed of the stream be ${y}$ km/h. Then the speed of the boat downstream = ${(x + y)}$ km/h, and the speed of the boat upstream = ${(x – y)}$ km/h Since ${Speed=(Distance)/(Time)}$. therefore ${Time=(Distance)/(Speed)}$ In First Case: Condition 1:when the boat goes 30 km upstream, let the time taken, in hour, be ${t_1}$ . Then ${t_1=30/(x-y)}$ Condition 2:Let ${t_2}$ be the time, in hours, taken by the boat to go 44 km downstream, then ${t_2=44/(x+y)}$ Condition 3: A boat goes 30 km upstream and 44 km downstream in 10 hours, Hence ${t_1+t_2=10}$ ${30/(x-y) + 44/(x+y) = 10}$ (1) In Second Case: Condition 1: Let ${t_3}$ be the time, in hours, taken by the boat to go 40 km upstream. ${t_3=40/(x-y)}$ Condition 2: Let ${t_4}$ be the time, in hours, taken by the boat to go 55 km downstream. ${t_4=55/(x+y)}$ Condition 3: In 13 hours, it can go 40 km upstream and 55 km down-stream. ${t_3+t_4=13}$ ${40/(x-y) + 55/(x+y) = 13}$ (2) Let ${1/(x-y) = u}$ and ${1/(x+y)=v}$, (3) Subsituting these in equations (1) and (2), pair of linear equations becomes: ${30u + 44v=10}$ (4) ${40u+55v=13}$ (5) Using any of the methods, we get; ${u=1/5}$ and ${v=1/11}$ Putting these values in equation (3); ${1/(x-y) = 1/5}$ and ${1/(x+y)=1/11}$ i.e., ${ x – y = 5}$ and ${x + y = 11}$ (6) Solving these equations: ${x=8}$ and ${y=3}$
Answer:Hence, the speed of the boat in still water is 8 km/h and the speed of the stream is 3 km/h.