## Linear Equations in Two Variables

### Class-10-Mathematics-1-Chapter-1-Maharashtra Board

### Solutions (Practice Set 1.3, 1.4, 1.5)

**Practice Set 1.3**

** Question 1.**

*1) Fill in the blanks with correct number*

\(\begin{vmatrix} 3 &2 \\ 4& 5 \end{vmatrix}\) *= 3 ×* * [ ] −** [ ] ×** 4 = [ ] −** 8 = [ ]*

\(\begin{vmatrix} 3 &2 \\ 4& 5 \end{vmatrix}\) *= 3 ×* * [ 5] −** [2 ] ×** 4 = [15 ] −** 8 = [ 7]*

**Question 2.**

*Find the values of following determinants.*

** (1) ** \(\begin{vmatrix} -1 &7 \\ 2& 4 \end{vmatrix}\)

\(\begin{vmatrix} -1 &7 \\ 2& 4 \end{vmatrix}\) *= (−**1) ×** 4 − **7 ×** 2 = −** 4 −** 14 = −**18*

*Answer is −**18*

** (2) ** \(\begin{vmatrix} 5 &3 \\ -7& 0 \end{vmatrix}\)

\(\begin{vmatrix} 5 &3 \\ -7& 0 \end{vmatrix}\) *= 5** ×** 0 − 3** ×** (−7) = 0 + 21** = 21*

*Answer is 21*

* *** (3) **\(\begin{vmatrix} \frac{7}{3} &\frac{5}{3} \\ \frac{3}{2}& \frac{1}{2} \end{vmatrix}\)

* * \(\begin{vmatrix} \frac{7}{3} &\frac{5}{3} \\ \frac{3}{2}& \frac{1}{2} \end{vmatrix}\)

= \(\frac{7}{3}\times \frac{1}{2}-\frac{5}{3}\times \frac{3}{2}\)=\(\frac{7}{6}-\frac{15}{6}=\frac{7-15}{6}\)

= \(-\frac{4}{3}\)

** Answer is**\(-\frac{4}{3}\)

* Question 3.*

* **Solve the following simultaneous equations using Cramer’s rule.*

*(1) 3**x −** 4**y **= 10 ; 4**x **+ 3**y **= 5 *

*3**x −** 4**y **= 10*

*4**x **+ 3**y **= 5 *

*In above given equations *

* a _{1 }= 3, b_{1 }= −*

*4, c*

_{1 }= 10* a _{2 }= 4, b_{2 }= 3, c_{2 }= 5 *

D = \(\begin{vmatrix} a_1 &b_1 \\ a_2& b_2 \end{vmatrix}\) ** ****=** \(\begin{vmatrix} 3 &-4 \\ 4& 3 \end{vmatrix}\)

*= 3 ×** 3 −**(−** 4) ×** 4 = 9 + 16 = 25*

D_{x} = \(\begin{vmatrix} c_1 &b_1 \\ c_2& b_2 \end{vmatrix}\) ** ****= **\(\begin{vmatrix} 10 &-4 \\ 5& 3 \end{vmatrix}\)

*= 10 × 3 −(− 4) × 5 = 30 + 20 **= 50*

D_{y} = \(\begin{vmatrix} a_1 &c_1 \\ a_2& c_2 \end{vmatrix}\) **= **\(\begin{vmatrix} 3 &10 \\ 4& 5 \end{vmatrix}\)

*= 3 × 5 − 10 × 4 ** = 15 − 40 = −** 25*

*By Cramers’s rule,*

\(x=\frac{D_x}{D} = \frac{50}{25}=2\)

and \(y=\frac{D_y}{D}= \frac{-25}{25}=-1\)

*Answer is (x, y) = (2, − 1)*

*(2) 4**x **+ 3**y −** 4 = 0 ; 6**x **= 8 − 5**y*

*Writing the equation in ax + by = c form*

* **4**x **+ 3**y =** 4*

* 6**x **+ 5**y** = 8 *

*In above given equations *

* a _{1 }= 4, b_{1 }= 3, c_{1 }= 4 *

* a _{2 }= 6, b_{2 }= 5, c_{2 }= 8*

D = \(\begin{vmatrix} a_1 &b_1 \\ a_2& b_2 \end{vmatrix}\) ** ****=** \(\begin{vmatrix} 4 &3 \\ 6& 5 \end{vmatrix}\) ** **

**=** *4 × 5 − 3 × 6 **= 20 −**18 = 2*

D_{x} = \(\begin{vmatrix} c_1 &b_1 \\ c_2& b_2 \end{vmatrix}\) ** ****=**\(\begin{vmatrix} 4 &3 \\ 6& 5 \end{vmatrix}\) ** **

**=** *4 × 5 − 3 × 8 ** = 20 **-** 24 = −** 4*

D_{y} = \(\begin{vmatrix} a_1 &c_1 \\ a_2& c_2 \end{vmatrix}\) ** ****=** \(\begin{vmatrix} 4 &3 \\ 6& 5 \end{vmatrix}\)

**=** 4 × 8 − 4 × 6 * = 32 −** 24 = 8*

*By Cramers’s rule,*

\(x=\frac{D_x}{D} = \frac{-4}{2}=-2\)

and \(y=\frac{D_y}{D}= \frac{8}{2}=4\)

*Answer is (x, y) = (− 2, 4)*

*(3) **x **+ 2**y **= − 1 ; 2**x − ** 3**y **= 12 *

*x **+ 2**y **= − 1 *

*2**x −** 3**y **= 12 *

*In above given equations *

* a _{1 }= 1, b_{1 }= 2, c_{1 }= −*

*1*

* a _{2 }= 2, b_{2 }= −*

*3, c*

_{2 }= 12D = \(\begin{vmatrix} a_1 &b_1 \\ a_2& b_2 \end{vmatrix}\) ** ****=** \(\begin{vmatrix} 1 &2 \\ 2& -3 \end{vmatrix}\)

**=** 1* × (− 3) − 2× 2** = −**3 − **4 = −**7*

D_{x} = \(\begin{vmatrix} c_1 &b_1 \\ c_2& b_2 \end{vmatrix}\)** ****= **\(\begin{vmatrix} -1 &2 \\ 12& -3 \end{vmatrix}\)

**=** *−*1* × (− 3) − 2× 12 **= 3 **−** 24 = −**21*

D_{y} = \(\begin{vmatrix} a_1 &c_1 \\ a_2& c_2 \end{vmatrix}\) **= **\(\begin{vmatrix} 1 &-1 \\ 2& 12 \end{vmatrix}\)

**=** 1 × 12 − 2 × −1*= 12 +2 = 14*

*By Cramers’s rule,*

\(x=\frac{D_x}{D} = \frac{-21}{-7}=3\)

and \(y=\frac{D_y}{D}= \frac{14}{-7}=-2\)

*Answer is (x, y) = (3, −2)*

*(4) 6**x −** 4**y **= −12 ; 8**x −** 3**y **= −2*

*6**x −** 4**y **= −12*

*8**x −** 3**y **= −2*

*In above given equations *

* a _{1 }= 6, b_{1 }= −4, c_{1 }= −*

*12*

* a _{2 }= 8, b_{2 }= −*

*3, c*

_{2 }= −2D = \(\begin{vmatrix} a_1 &b_1 \\ a_2& b_2 \end{vmatrix}\) ** ****=** \(\begin{vmatrix} 6 &-4 \\ 8& -3 \end{vmatrix}\)

**=** 6* × (− 3) − (−4) × 8 *** ****=** * −**18 + 32 = 14*

D_{x} = \(\begin{vmatrix} c_1 &b_1 \\ c_2& b_2 \end{vmatrix}\) ** ****= **\(\begin{vmatrix} -12 &-4 \\ -2& -3 \end{vmatrix}\)

**=** (*− 12)** × (− 3) − (−4) × (− 2) **= 36 −** 8 = 28*

D_{y} = \(\begin{vmatrix} a_1 &c_1 \\ a_2& c_2 \end{vmatrix}\) **= **\(\begin{vmatrix} 6 &-12 \\ 8& -2 \end{vmatrix}\)

**=** 6* × (− 2) − (−12) × 8 ** = −**12 + 96 = 84*

*By Cramers’s rule,*

\(x=\frac{D_x}{D} = \frac{28}{14}=2\)

and \(y=\frac{D_y}{D}= \frac{84}{14}=6\)

*Answer is (x, y) = (2, 6)*

*(5) 4**m **+ 6**n **= 54 ; 3**m **+ 2**n **= 28 *

*4**m **+ 6**n **= 54*

*3**m **+ 2**n **= 28 *

*In above given equations *

* a _{1 }= 4, b_{1 }= 6, c_{1 }= 54 *

* a _{2 }= 3, b_{2 }= 2, c_{2 }= 28 *

D = \(\begin{vmatrix} a_1 &b_1 \\ a_2& b_2 \end{vmatrix}\) ** ****=** \(\begin{vmatrix} 4 &6 \\ 3& 2 \end{vmatrix}\)

**=** 4* × 2 − 6 × 3 *** ****=** * 8 − **18 = **− **10*

D_{m} = \(\begin{vmatrix} c_1 &b_1 \\ c_2& b_2 \end{vmatrix}\) **= **\(\begin{vmatrix} 54 &6 \\ 28& 2 \end{vmatrix}\)

**=** 54* × 2 − 6 × 28 **= 108 −** 168 = −**60*

D_{n} = \(\begin{vmatrix} a_1 &c_1 \\ a_2& c_2 \end{vmatrix}\) ** ****= **\(\begin{vmatrix} 4 &54 \\ 3& 28 \end{vmatrix}\)

**=** 4* × 2 8 − 54 × 3 *** ****= ***112 −**162 = −**50*

*By Cramers’s rule,*

\(x=\frac{D_m}{D} = \frac{-60}{-10}=6\)

and \(y=\frac{D_n}{D}= \frac{-50}{-10}=5\)

*Answer is (m, n) = (6, 5)*

*(6) 2**x **+ 3**y **= 2 ; **x *** –** \(\frac{y}{2}=\frac{1}{2}\)

*2**x **+ 3**y **= 2*

* **x **– *\(\frac{y}{2}=\frac{1}{2}\)

*In above given equations *

* a _{1 }= 2, b_{1 }= 3, c_{1 }= 2 *

* a _{2 }= 1, b_{2 }= *\(-\frac{1}{2}\)

*, c*\(\frac{1}{2}\)

_{2 }=D = \(\begin{vmatrix} a_1 &b_1 \\ a_2& b_2 \end{vmatrix}\) ** ****=** \(\begin{vmatrix} 2 &3 \\ 1& -\frac{1}{2} \end{vmatrix}\)

**=** 2* ×* (\(-\frac{1}{2}|)) *− 3 × 1 *** ****=*** –**1 –** 3 = –**4*

D_{x} = \(\begin{vmatrix} c_1 &b_1 \\ c_2& b_2 \end{vmatrix}\) **= **\(\begin{vmatrix} 2 &3 \\ \frac{1}{2}& -\frac{1}{2} \end{vmatrix}\)

**=** 2* × *(\(-\frac{1}{2}\)) *− 3 × *\(\frac{1}{2}\) ** ****= ***−1−* \(\frac{3}{2}\) = *−* \(\frac{5}{2}\)

D_{y} = \(\begin{vmatrix} a_1 &c_1 \\ a_2& c_2 \end{vmatrix}\) ** ****= **\(\begin{vmatrix} 2 &2 \\ 1& \frac{1}{2} \end{vmatrix}\)

**=** 2* × *(\(\frac{1}{2}\))* − 2 × 1 ***= *** 1 −** 2 = −**1*

*By Cramers’s rule,*

\(x=\frac{D_x}{D} = \frac{-\frac{5}{2}}{-4}=\frac{5}{8}\)

and \(y=\frac{D_y}{D}= \frac{-1}{-4}=\frac{1}{4}\)

** Answer is (x, y) =** (\(\frac{5}{8}\), \(\frac{1}{4}\))

**Practice Set 1.4**

*Question 1.*

*Solve the following simultaneous equations.*

** (1) ** \(\frac{2}{x}-\frac{3}{y}\) = 15, \(\frac{8}{x}+\frac{5}{y}\) = 77

\(\frac{2}{x}-\frac{3}{y}\) = 15 * **…….. **(1)*

\(\frac{8}{x}+\frac{5}{y}\) = 77 *…….. **(2)*

*Substituting a for *\(\frac{1}{x}\) * and b for *\(\frac{1}{y}\) * the given equations become,*

*2a − 3b = 15 …….(3)*

*8a + 5b = 77 ……(4)*

*Multiplying equation (3) by 4,*

*8a − 12b = 60 ……(5)*

*Subtracting equation (5) from equation (4),*

\(\begin{array}{rrrrrrr}

& 8a & + & 5b & = & 77 &\text{…..(4)}\\

- & & & & & & \\

&8a & - & 12b & = & 60 &\text{…..(5)} \\

-& & +& & &-& \\ \hline

& & & 17b & = & 17 & \\

\end{array}\)

*∴** b = 1*

*Substituting b = 1 in equation (3),*

*2a −3(1) = 15, ∴** 2a – 3 = 15, ∴** 2a = 15 + 3*

*2a = 18, ∴ **a=9*

*Resubstituting the values of a and b,*

*a = *\(\frac{1}{x}\) * **= 9 and b = *\(\frac{1}{y}\) * = 1*

*∴ x = *\(\frac{1}{9}\) * , and y = 1*

** Answer is (x, y) = (** \(\frac{1}{9}\),

*1)*

** (2) ** \(\frac{10}{x+y}+\frac{2}{x-y}\) = 4, \(\frac{15}{x-y}-\frac{5}{x-y}\) =

*–*2

\(\frac{10}{x+y}+\frac{2}{x-y}\) = 4 *…….. **(1)*

\(\frac{15}{x-y}-\frac{5}{x-y}\) = –2 *…….. **(2)*

*Substituting a for *\(\frac{1}{x+y}\) * and b for *\(\frac{1}{x-y}\) * the given equations become,*

*10a + 2b = 4 …….(3)*

*15a – 5b = –2 ……(4)*

*Multiplying equation (3) by 5,*

*50a + 10b = 20 ……(5)*

*Multiplying equation (4) by 2,*

*30a – 10b = –4 ……(6)*

*Adding equation (5) and (6),*

\(\begin{array}{rrrrrrr}

& 50a & + & 10b & = & 20 &\text{…..(5)}\\

+ & & & & & & \\

&30a & - & 10b & = & -4 &\text{…..(6)} \\

& & & & && \\ \hline

&80a & & & = & 16 & \\

\end{array}\)

* ∴ **a = *\(\frac{16}{80}\) = \(\frac{1}{5}\)

*Substituting a = *\(\frac{1}{5}\) *in equation (3),*

\(10(\frac{1}{5})\) *+ 2b = 4, ∴ **2 + 2b = 4, ∴** 2b = 4 – 2 = 2*

*∴ b = 1 *

*Resubstituting the values of a and b,*

*a = *\(\frac{1}{x+y}=\frac{1}{5}\) * ** and b = *\(\frac{1}{x-y}\) *= 1*

*∴ x + y = 5 ….(7) ** , and x **—** y = 1 …..(8)*

*Adding equation (5) and (6),*

\(\begin{array}{rrrrrrr}

& x & + & y & = & 5 &\text{…..(7)}\\

+ & & & & & & \\

&x & - & y & = & 1 &\text{…..(8)} \\

& & & & && \\ \hline

&2x & & & = & 6 & \\

\end{array}\)

*∴ x = 6/2 = 3*

*Substituting x = 3 in equ. (7),*

*3 + y = 5, ∴** y = 5 – 3 = 2*

* **Answer is (x, y) = (3, 2**)*

** (3) ** \(\frac{27}{x-2}+\frac{31}{y+3}\) = 85, \(\frac{31}{x-2}-\frac{27}{y+3}\) = 89

\(\frac{27}{x-2}+\frac{31}{y+3}\) = 85 *…….. **(1)*

\(\frac{31}{x-2}-\frac{27}{y+3}\) = 89* …….. **(2)*

*Substituting a for * \(\frac{1}{x-2}\)* and b for * \(\frac{1}{y+3}\)* the given equations become, *

*27a + 31b = 85 …….(3)*

*31a – 27b = 89 ……(4)*

*Adding equation (3) and (4),*

\(\begin{array}{rrrrrrr}

& 27a & + & 31b & = & 85 &\text{…..(3)}\\

+ & & & & & & \\

&31a & - & 27b & = & 89 &\text{…..(4)} \\

& & & & && \\ \hline

&58a &+ &58b & = & 174 & \\

\end{array}\)

* ** ∴ ** a + b = 3 (Dividing both sides by 58)….. (5)*

* **Subtracting equation (3) from equation (4),*

\(\begin{array}{rrrrrrr}

& 31a & - & 27b & = & 89 &\text{…..(3)}\\

- & & & & & & \\

&27a & + & 31b & = & 85 &\text{…..(4)} \\

&- &- & & &-& \\ \hline

&4a &- &4b & = & 4 & \\

\end{array}\)

* ∴ **a - b = 1 (Dividing both sides by 4) … (6)*

* **Adding equation (5) and (6),*

\(\begin{array}{rrrrrrr}

& a & + & b & = & 3 &\text{…..(5)}\\

+ & & & & & & \\

&a & - & b & = & 1 &\text{…..(6)} \\

& & & & && \\ \hline

&2a & & & = & 4 & \\

\end{array}\)

*∴ a = 2*

* **Substituting a = 2 in equation (5),*

*2 + b = 3, ∴** b = 3 – 2 , **∴** b = 1 *

*Resubstituting the values of a and b,*

*a = *\(\frac{1}{x-2}\) = 2

*∴ 2(x – 2) = 1,* *∴ **2**x – 4 = 1, ∴** 2**x = 1+ 4 =5 ∴**x = 5/2*

* b = *\(\frac{1}{y+3}\) * = 1*

*∴ y + 3 = 1, **∴** y = 1 – 3 = –2 , **∴ **y = – 2 *

*Answer is (x, y) = (**5/2, – 2**)*

** (4) ** \(\frac{1}{3x+y}+\frac{1}{3x-y}\) = \(\frac{3}{4}\), \(\frac{1}{2(3x+y)}-\frac{1}{2(3x-y}\) = \(-\frac{1}{8}\)

\(\frac{1}{3x+y}+\frac{1}{3x-y}\) = \(\frac{3}{4}\) *…….. **(1)*

\(\frac{1}{2(3x+y)}-\frac{1}{2(3x-y}\) = \(-\frac{1}{8}\) *…….. **(2)*

*Substituting a for *\(\frac{1}{3x+y}\) * and b for *\(\frac{1}{3x-y}\) *the given equations become, *

*a + b = *\(\frac{3}{4}\) * …….(3)*

\(\frac{a}{2}-\frac{b}{2}\) = \(-\frac{1}{8}\)* ……(4)*

*Multiplying equation (3) by 4*

*4a + 4b = 3 …….(5)*

*Multiplying equation (4) by 8*

*4a – 4b = – 1 …….(6)*

*Adding equation (5) and (6),*

\(\begin{array}{rrrrrrr}

& 4a & + & 3b & = & 3 &\text{…..(3)}\\

+ & & & & & & \\

&4a & - & 4b & = & -1 &\text{…..(4)} \\

& & & & && \\ \hline

&8a & & & = & 2 & \\

\end{array}\)

*∴ a = *\(\frac{1}{4}\)

*Substituting a =*\(\frac{1}{4}\) *in equation (5)*

*4(¼) + 4b = 3, ∴ **1 + 4b = 3, ∴ **4b = 3 – 1 = 2 *

*∴b = *\(\frac{1}{2}\)

*Resubstituting the values of a and b,*

*a = *\(\frac{1}{3x+y}\) = \(\frac{1}{4}\)

*∴ 3x + y = 4 …..(7) *

*And ** *

* b = *\(\frac{1}{3x-y}\) =\(\frac{1}{2}\)

*∴ 3x – y = 2 …….(8)*

*Adding equation (7) and (8),*

\(\begin{array}{rrrrrrr}

& 3x & + & y & = & 4 &\text{…..(7)}\\

+ & & & & & & \\

&3x & - & y & = & 2 &\text{…..(8)} \\

& & & & && \\ \hline

&6x & & & = & 6 & \\

\end{array}\)

*∴ x = 1*

*Substituting x =1 in equation 7 *

*∴ 3 + y = 4 ∴** y = 4 – 3 = 1*

*Answer is (x, y) = (1, 1**)*

**Practice Set 1.5**

*(1) Two numbers differ by 3. The sum of twice the smaller number and thrice the greater number is 19. Find the numbers.*

*Let the greater number be x and the smaller number be y.*

*From the ﬁrst condition,*

*x – y = 3 …..(1)*

*Thrice the greater number = 3x.*

*Twice the smaller number = 2y.*

*From the second condition,*

*3x + 2y = 19*

*Multiplying equation (1) by 2,*

*2x – 2y = 6 …..(3)*

*Adding equations (2) and (3),*

\(\begin{array}{rrrrrrr}

& 3x & + & 2y & = & 19 &\text{…..(2)}\\

+ & & & & & & \\

&2x & - & 2y & = & 6 &\text{…..(3)} \\

& & & & && \\ \hline

&5x & & & = & 25 & \\

\end{array}\)

* ∴ **x = 5*

*Substituting x = 5 in equation (1),*

*5 – y = 3 – y = 3 – 5 –y = –2 ∴* *y = 2*

*The required numbers are 5 and 2 is the answer.*

*(2) Complete the following.*

*The opposite sides of a rectangle are of equal lengths.*

* 2x + y + 8 = 4x – y*

*∴ 2x – 4x + y + y = – 8*

* ∴ **– **2x + 2y = – 8*

*∴ – x + y = – 4 (Dividing both the sides by 2) …..(1)*

*Also, x + 4 = 2y*

*∴ x – 2y = – 4 ……..(2)*

*Adding equations (1) and (2),*

\(\begin{array}{rrrrrrr}

& -x & + & y & = & -4 &\text{…..(1)}\\

+ & & & & & & \\

&x & - & 2y & = & -4 &\text{…..(2)} \\

& & & & && \\ \hline

& & & -y& = & -8 & \\

\end{array}\)

*∴ y = 8*

*Substituting y = 8 in equation (2),*

*x – 2(8) = – 4, ∴** x –16 = – 4 , ∴** x = – 4 + 16 = 12*

* ∴** x = 12*

*The length of the rectangle = 4x – y = 4(12) – 8 = 48 – 8 = 40*

*The breadth of the rectangle = 2y = 2 ×** 8 = 16*

*The area of a rectangle = l ×** b*

*= 40 ×** 16 = 640 sq. units*

*The perimeter of the rectangle = 2(l + b)*

*= 2(40 + 16)*

*= 2 ×** 56= 112 units*

*Answer is x = 12; y = 8. The area of the rectangle is 640 sq. units and its perimeter is 112 units**.*

* (3) The sum of father’s age and twice the age of his son is 70. If we double the age of the father and add it to the age of his son the sum is 95. Find their present ages.*

*Let the father’s present age be x years and the son’s present age be y years.*

*From the first condition,*

*x + 2y = 70 …… (1)*

*Double the age of father = 2x.*

*From the second condition,*

*2x + y = 95 …..(2)*

*Multiplying equation (1) by 2,*

*2x + 4y = 140 ……(3)*

*Subtracting equation (2) from equation (3),*

\(\begin{array}{rrrrrrr}

& 2x &+ & 4y & = & 140 &\text{…..(3)}\\

- & & & & & & \\

& 2x & + & y & = & 95 &\text{…..(2)} \\

-& &- & & &-& \\ \hline

& & & 3y& = & 45 & \\

\end{array}\)

*∴ y = 15*

*Substituting y = 15 in equation (1),*

*x + 2(15) = 70, ∴ **x + 30 = 70*

*∴ x = 70 – 30 , ∴** x = 40*

*Answer is the father’s present age is 40 years and the son’s present age is 15 years.*

*(4) The denominator of a fraction is 4 more than twice its numerator. Denominator becomes 12 times the numerator, if both the numerator and the denominator are reduced by 6. Find the fraction.*

*Let the numerator of the fraction be x and its denominator be y.*

*Twice the numerator = 2x.*

*From the ﬁrst condition,*

*y = 2x + 4, ** ∴ **2x + 4 = y, ∴ **2x –** y = –** 4 ...(1)*

*If 6 is subtracted both from the numerator and the denominator, then the numerator becomes (x –** 6) and the denominator becomes ( y –** 6).*

*From the second condition, *

*y – 6 = 12(x – 6)*

*∴ y – 6 = 12x –** 72*

* 12x – 72 = y –** 6*

* 12x – y = **-** 6 + 72*

* 12x –** y = 66 ……(2)*

*Subtracting equation (1) from equation (2),*

\(\begin{array}{rrrrrrr}

& 12x & - & y & = & 66 &\text{…..(2)}\\

- & & & & & & \\

& 2x & - & y & = & -4 &\text{…..(1)} \\

-& &+ & & &+ & \\ \hline

& 10x & & & = & 70 & \\

\end{array}\)

*∴ x = 7*

*Substituting x = 7 in equation (1),*

*2(7) – y = –** 4*

* 14 – y = – 4*

* – y = – 4 – 14*

* – y = –**18, ∴* *y = 18*

*The numerator, x is 7 and the denominator, y is 18*

** Answer is- The required fraction is **\(\frac{7}{18}\)

*(5) Two types of boxes A, B are to be placed in a truck having capacity of 10 tons.*

*When 150 boxes of type A and 100 boxes of type B are loaded in the truck, it weighes 10 tons. But when 260 boxes of type A are loaded in the truck, it can still accommodate 40 boxes of type B, so that it is fully loaded. Find the weight of each type of box.*

*Let the weight of a box of type A be x ton and that of a box of type B be y ton.*

*The weight of 150 boxes of type A = 150x ton.*

*The weight of 100 boxes of type B = 100 y ton.*

*From the first condition,*

*150x + 100y = 10*

* 15x + 10y = 1 (Dividing both the sides by 10) …… (1)*

*Similarly, from the second condition,*

*260x + 40y = 10*

* 26x + 4y = 1 (Dividing both the sides by 10) …… (2)*

*Multiplying equation (1) by 2 and equation (2) by 5,*

*30x + 20y = 2 ….(3)*

*130x + 20y = 5 ……(4)*

*Subtracting equation (3) from equation (4),*

\(\begin{array}{rrrrrrr}

& 130x & + & 20y & = & 5 &\text{…..(4)}\\

- & & & & & & \\

& 30x & + & 20y & = & 2 &\text{…..(3)} \\

-& &- & & &- & \\ \hline

& 100x & & & = & 3 & \\

\end{array}\)

*∴ x = *\(\frac{3}{100}\)

*Substituting x = \(\frac{3}{100}\)* * in equation (1),*

*15*(\(\frac{3}{100}\))* + 10y = 1*

\(\frac{45}{100}\)* + 10y = 1*

*10y = 1- *\(\frac{45}{100}\)

*∴ 10y = * \(\frac{55}{100}\) *= *

*∴ y = *\(\frac{45}{1000}\) * ...(Dividing both the sides by 10)*

** Answer is - The weight of a box of type A is** \(\frac{3}{100}\)

*ton*

* i.e. ** ton = *\(\frac{3}{100}\) * ×** 1000 kg = 30 kg and *

** the weight of a box of type B is **\(\frac{55}{1000}\)

*ton*

*i.e. ** ton = *\(\frac{55}{1000}\) *×* *1000 kg = 55kg*

*(6) Out of 1900 km, Vishal travelled some distance by bus and some by aeroplane. Bus travels with average speed 60 km/hr and the average speed of aeroplane is 700 km/hr. It takes 5 hours to complete the journey. Find the distance, Vishal travelled by bus.*

*Let Vishal travelled x km by bus and y km by plane.*

*Time = (distance)/(speed)*

*Time taken by Vishal while travelling by bus*

*= *\(\frac{x}{60}\)*h*

*Time taken by Vishal while travelling by plane*

*= *\(\frac{y}{700}\)*h*

*From the given conditions,*

*x + y = 1900 ……(1)*

*and *\(\frac{x}{60}+\frac{y}{700}\) *= 5 ….(2)*

*Multiplying equation (2) by 2100, the LCM of 60 and 700, *

\(\frac{x}{60}×2100+\frac{y}{700}×2100\) *= 5 × **2100*

*∴ 35x + 3y = 10500 …….(3)*

*Multiplying equation (1) by 3,*

*3x + 3y = 5700 …..(4)*

*Subtracting equation (4) from equation (3),*

\(\begin{array}{rrrrrrr}

& 35x & + & 3y & = & 10500 &\text{…..(3)}\\

- & & & & & & \\

& 3x & + & 3y & = & 5700 &\text{…..(4)} \\

-& &- & & &- & \\ \hline

& 32x & & & = & 4800 & \\

\end{array}\)

* ∴ **x = *\(\frac{4800}{32}\)* = 150*

* ∴ **x = 150*

*Answer is- Vishal travelled 150 km by bus.*

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