SSC 10th Maharashtra Board: Mathematics Part 1 (Practice Set 1.4)
Question 1
Solve the following simultaneous equations:
$$\frac{2}{x} - \frac{3}{y} = 15$$
$$\frac{8}{x} + \frac{5}{y} = 77$$
Solution:
Let $\frac{1}{x} = m$ and $\frac{1}{y} = n$. Substituting these values in the given equations, we get:
$$2m - 3n = 15 \quad \text{--- (I)}$$
$$8m + 5n = 77 \quad \text{--- (II)}$$
Multiplying equation (I) by $4$, we get:
$$8m - 12n = 60 \quad \text{--- (III)}$$
Subtracting equation (III) from equation (II):
$$(8m + 5n) - (8m - 12n) = 77 - 60$$
$$17n = 17 \implies n = 1$$
Substitute $n = 1$ in equation (I):
$$2m - 3(1) = 15 \implies 2m - 3 = 15 \implies 2m = 18 \implies m = 9$$
Now, substituting the values of $m$ and $n$ back to $x$ and $y$:
$m = \frac{1}{x} \implies 9 = \frac{1}{x} \implies x = \frac{1}{9}$
$n = \frac{1}{y} \implies 1 = \frac{1}{y} \implies y = 1$
The solution is $(x, y) = \left(\frac{1}{9}, 1\right)$.
Question 2
Solve the following simultaneous equations:
$$\frac{10}{x+y} + \frac{2}{x-y} = 4$$
$$\frac{15}{x+y} - \frac{5}{x-y} = -2$$
Solution:
Let $\frac{1}{x+y} = a$ and $\frac{1}{x-y} = b$. The equations become:
$$10a + 2b = 4 \implies 5a + b = 2 \quad \text{--- (I) (Dividing by 2)}$$
$$15a - 5b = -2 \quad \text{--- (II)}$$
Multiplying equation (I) by $5$, we get:
$$25a + 5b = 10 \quad \text{--- (III)}$$
Adding equation (II) and equation (III):
$$(15a - 5b) + (25a + 5b) = -2 + 10$$
$$40a = 8 \implies a = \frac{8}{40} \implies a = \frac{1}{5}$$
Substitute $a = \frac{1}{5}$ in equation (I):
$$5\left(\frac{1}{5}\right) + b = 2 \implies 1 + b = 2 \implies b = 1$$
Resubstituting the values of $a$ and $b$:
$a = \frac{1}{x+y} \implies \frac{1}{5} = \frac{1}{x+y} \implies x + y = 5 \quad \text{--- (IV)}$
$b = \frac{1}{x-y} \implies 1 = \frac{1}{x-y} \implies x - y = 1 \quad \text{--- (V)}$
Adding equations (IV) and (V):
$$(x + y) + (x - y) = 5 + 1 \implies 2x = 6 \implies x = 3$$
Substitute $x = 3$ in equation (IV):
$$3 + y = 5 \implies y = 2$$
The solution is $(x, y) = (3, 2)$.
Question 3
Solve the following simultaneous equations:
$$\frac{27}{x-2} + \frac{31}{y+3} = 85$$
$$\frac{31}{x-2} + \frac{27}{y+3} = 89$$
Solution:
Let $\frac{1}{x-2} = p$ and $\frac{1}{y+3} = q$. The equations become:
$$27p + 31q = 85 \quad \text{--- (I)}$$
$$31p + 27q = 89 \quad \text{--- (II)}$$
Adding equations (I) and (II):
$$58p + 58q = 174 \implies p + q = 3 \quad \text{--- (III) (Dividing by 58)}$$
Subtracting equation (I) from equation (II):
$$(31p + 27q) - (27p + 31q) = 89 - 85$$
$$4p - 4q = 4 \implies p - q = 1 \quad \text{--- (IV) (Dividing by 4)}$$
Adding equations (III) and (IV):
$$(p + q) + (p - q) = 3 + 1 \implies 2p = 4 \implies p = 2$$
Substitute $p = 2$ in equation (III):
$$2 + q = 3 \implies q = 1$$
Resubstituting the values of $p$ and $q$:
$p = \frac{1}{x-2} \implies 2 = \frac{1}{x-2} \implies 2(x - 2) = 1 \implies 2x - 4 = 1 \implies 2x = 5 \implies x = \frac{5}{2}$
$q = \frac{1}{y+3} \implies 1 = \frac{1}{y+3} \implies y + 3 = 1 \implies y = -2$
The solution is $(x, y) = \left(\frac{5}{2}, -2\right)$.
Question 4
Solve the following simultaneous equations:
$$\frac{1}{3x+y} + \frac{1}{3x-y} = \frac{3}{4}$$
$$\frac{1}{2(3x+y)} - \frac{1}{2(3x-y)} = -\frac{1}{8}$$
Solution:
Let $\frac{1}{3x+y} = u$ and $\frac{1}{3x-y} = v$. The equations become:
$$u + v = \frac{3}{4} \implies 4u + 4v = 3 \quad \text{--- (I)}$$
$$\frac{u}{2} - \frac{v}{2} = -\frac{1}{8} \implies u - v = -\frac{1}{4} \implies 4u - 4v = -1 \quad \text{--- (II)}$$
Adding equations (I) and (II):
$$(4u + 4v) + (4u - 4v) = 3 - 1 \implies 8u = 2 \implies u = \frac{1}{4}$$
Substitute $u = \frac{1}{4}$ into the first simplified form $u + v = \frac{3}{4}$:
$$\frac{1}{4} + v = \frac{3}{4} \implies v = \frac{3}{4} - \frac{1}{4} = \frac{2}{4} = \frac{1}{2}$$
Resubstituting the values of $u$ and $v$:
$u = \frac{1}{3x+y} \implies \frac{1}{4} = \frac{1}{3x+y} \implies 3x + y = 4 \quad \text{--- (III)}$
$v = \frac{1}{3x-y} \implies \frac{1}{2} = \frac{1}{3x-y} \implies 3x - y = 2 \quad \text{--- (IV)}$
Adding equations (III) and (IV):
$$(3x + y) + (3x - y) = 4 + 2 \implies 6x = 6 \implies x = 1$$
Substitute $x = 1$ in equation (III):
$$3(1) + y = 4 \implies 3 + y = 4 \implies y = 1$$
The solution is $(x, y) = (1, 1)$.
