Practice Set 1.4 Algebra 10th Standard Maths Part 1 Chapter 1 Linear Equations in Two Variables Solutions Maharashtra Board

Practice Set 1.4 Solutions

The given simultaneous equations are: $\frac{2}{x}-\frac{3}{y}=15$ ... (i) and $\frac{8}{x}+\frac{5}{y}=77$ ... (ii)
Let $\frac{1}{x}=p$ and $\frac{1}{y}=q$
Equations (i) and (ii) become: $2p-3q=15$ ... (iii) and $8p+5q=77$ ... (iv)
Multiplying equation (iii) by 4, we get $8p-12q=60$ ... (v)
Subtracting equation (v) from (iv) gives: $17q = 17 \Rightarrow q = 1$
Substituting $q=1$ in equation (iii): $2p-3(1)=15 \Rightarrow 2p=18 \Rightarrow p=9$
Resubstituting the values of p and q: $9=\frac{1}{x} \Rightarrow x=\frac{1}{9}$ and $1=\frac{1}{y} \Rightarrow y=1$
$(x,y)=(\frac{1}{9},1)$ is the solution.

The given simultaneous equations are: $\frac{10}{x+y}+\frac{2}{x-y}=4$ ... (i) and $\frac{15}{x+y}-\frac{5}{x-y}=-2$ ... (ii)
Let $\frac{1}{x+y}=p$ and $\frac{1}{x-y}=q$
Equations become: $10p+2q=4 \Rightarrow 5p+q=2$ ... (iii) and $15p-5q=-2$ ... (iv)
Multiplying equation (iii) by 5: $25p+5q=10$ ... (v)
Adding (iv) and (v): $40p=8 \Rightarrow p=\frac{1}{5}$
Substituting $p=\frac{1}{5}$ in equation (iii): $5(\frac{1}{5})+q=2 \Rightarrow 1+q=2 \Rightarrow q=1$
Resubstituting values: $\frac{1}{5}=\frac{1}{x+y} \Rightarrow x+y=5$ ... (vi) and $1=\frac{1}{x-y} \Rightarrow x-y=1$ ... (vii)
Adding (vi) and (vii): $2x=6 \Rightarrow x=3$
Substituting $x=3$ in (vi): $3+y=5 \Rightarrow y=2$
$(x,y)=(3,2)$ is the solution.

Let $\frac{1}{x-2}=p$ and $\frac{1}{y+3}=q$
Equations become: $27p+31q=85$ ... (iii) and $31p+27q=89$ ... (iv)
Adding (iii) and (iv): $58p+58q=174 \Rightarrow p+q=3$ ... (v)
Subtracting (iv) from (iii): $-4p+4q=-4 \Rightarrow p-q=1$ ... (vi)
Adding (v) and (vi): $2p=4 \Rightarrow p=2$
Substituting $p=2$ in (v): $2+q=3 \Rightarrow q=1$
Resubstituting values: $2=\frac{1}{x-2} \Rightarrow 2x-4=1 \Rightarrow x=\frac{5}{2}$ and $1=\frac{1}{y+3} \Rightarrow y+3=1 \Rightarrow y=-2$
$(x,y)=(\frac{5}{2},-2)$ is the solution.

Let $\frac{1}{3x+y}=p$ and $\frac{1}{3x-y}=q$
Equations become: $p+q=\frac{3}{4}$ ... (iii) and $\frac{1}{2}p-\frac{1}{2}q=-\frac{1}{8}$ ... (iv)
Multiplying (iv) by 2: $p-q=-\frac{1}{4}$ ... (v)
Adding (iii) and (v): $2p = \frac{2}{4} \Rightarrow p=\frac{1}{4}$
Substituting $p=\frac{1}{4}$ in (iii): $\frac{1}{4}+q=\frac{3}{4} \Rightarrow q=\frac{1}{2}$
Resubstituting values: $\frac{1}{4}=\frac{1}{3x+y} \Rightarrow 3x+y=4$ ... (vi) and $\frac{1}{2}=\frac{1}{3x-y} \Rightarrow 3x-y=2$ ... (vii)
Adding (vi) and (vii): $6x=6 \Rightarrow x=1$
Substituting $x=1$ in (vi): $3(1)+y=4 \Rightarrow y=1$
$(x,y)=(1,1)$ is the solution.

Complete the following table.

In the above table the equations are not linear. Can you convert the equations into linear equations?

Yes, the above given simultaneous equations can be converted to a pair of linear equations by making suitable substitutions.

Steps for solving equations reducible to a pair of linear equations:

Select suitable variables other than those which are in the equations.
Replace the given variables with new variables such that the given equations become linear equations in two variables.
Solve the new simultaneous equations and find the values of the new variables.
By resubstituting the value(s) of the new variables, find the replaced variables which are to be determined.

To solve given equations fill the below boxes suitably.

Equations: $\frac{5}{x-1}+\frac{1}{y-2}=2$; $\frac{6}{x-1}-\frac{3}{y-2}=1$

Replacing $(\frac{1}{x-1})$ by $m$, $(\frac{1}{y-2})$ by $n$.
New equations: $5m+n=2$ and $6m-3n=1$
On solving: $m=\boxed{\frac{1}{3}}$, $n=\boxed{\frac{1}{3}}$
Replacing m, n by their original values: $\frac{1}{x-1}=\frac{1}{3}$ and $\frac{1}{y-2}=\frac{1}{3}$
On solving: $x-1=3$, $y-2=3 \Rightarrow x=\boxed{4}, y=\boxed{5}$
$(x,y)=(4,5)$ is the solution.

Solve equations by graphical method.

ii. $\frac{4}{x-y}+\frac{1}{x+y}=3$; $\frac{2}{x-y}-\frac{3}{x+y}=5$

Let $\frac{1}{x-y}=p$ and $\frac{1}{x+y}=q$
Equations become: $4p+q=3 \Rightarrow q=3-4p$ and $2p-3q=5 \Rightarrow q=\frac{2p-5}{3}$

Table for $4p+q=3$:

Table for $2p-3q=5$:

The two lines intersect at point (1,-1). Therefore $p=1$ and $q=-1$.

Resubstituting values:

Table for $x-y=1$:

Table for $x+y=-1$:

The two lines intersect at point (0,-1). Therefore $(x,y)=(0,-1)$ is the solution.