Solve the following simultaneous equations. 7 2 x + 1 + 13 y + 2 = 27 ; 13 2 x + 1 + 7 y + 2 = 33

Given: 7 2 x + 1 + 13 y + 2 = 27 ; 13 2 x + 1 + 7 y + 2 = 33 Let 1 2 x + 1 = u and 1 y + 2 = v 7 u + 13 v = 27 … … ( i ) 13 u + 7 v = 33 … … ( ii ) Add eq. i & ii 20 u + 20 v = 60 ⇒ u + v = 3 … … ( iii ) Now eq. ii - i 6 u - 6 v = 6 ⇒ u - v = 1 … … ( iv ) On adding eq. iii & iv 2 u = 4 ⇒ u = 2 Putting the value of u in iv 2 - v = 1 ⇒ v = 1 Since, 1 2 x + 1 = u = 2 ⇒ 2 x + 1 = 1 2 ⇒ 2 x = 1 2 - 1 ⇒ 2 x = - 1 2 ⇒ x = - 1 4 and 1 y + 2 = v = 1 ⇒ y + 2 = 1 ⇒ y = - 1 ∴ ( x , y ) = - 1 4 , - 1

Solve the following simultaneous equations. 7 2 x + 1 + 13 y + 2 = 27 ; 13 2 x + 1 + 7 y + 2 = 33

Given: 7 2 x + 1 + 13 y + 2 = 27 ; 13 2 x + 1 + 7 y + 2 = 33 Let 1 2 x + 1 = u and 1 y + 2 = v 7 u + 13 v = 27 … … ( i ) 13 u + 7 v = 33 … … ( ii ) Add eq. i & ii 20 u + 20 v = 60 ⇒ u + v = 3 … … ( iii ) Now eq. ii - i 6 u - 6 v = 6 ⇒ u - v = 1 … … ( iv ) On adding eq. iii & iv 2 u = 4 ⇒ u = 2 Putting the value of u in iv 2 - v = 1 ⇒ v = 1 Since, 1 2 x + 1 = u = 2 ⇒ 2 x + 1 = 1 2 ⇒ 2 x = 1 2 - 1 ⇒ 2 x = - 1 2 ⇒ x = - 1 4 and 1 y + 2 = v = 1 ⇒ y + 2 = 1 ⇒ y = - 1 ∴ ( x , y ) = - 1 4 , - 1

Solve the following simultaneous equations. 148 x + 231 y = 527 x y ; 231 x + 148 y = 610 x y

Given: 148 x + 231 y = 527 x y ; 231 x + 148 y = 610 x y Multiply both the equations by x y 148 y + 231 x = 527 . . . . . . ( i ) 231 y + 148 x = 610 . . . . . . ( ii ) Adding i and ii 379 y + 379 x = 1137 ⇒ x + y = 3 . . . . . . iii Again, ii - i 83 y - 83 x = 83 ⇒ y - x = 1 . . . . . . iv Now adding iii & iv 2 y = 4 ⇒ y = 2 Putting the value of y in iv 2 - x = 1 ⇒ x = 1 ∴ ( x , y ) = ( 1 , 2 )

Solve the following simultaneous equations. 7 x - 2 y x y = 5 ; 8 x + 7 y x y = 15

Given: 7 x - 2 y x y = 5 ; 8 x + 7 y x y = 15 7 x - 2 y x y = 5 ⇒ 7 x x y - 2 y x y = 5 ⇒ 7 y - 2 x = 5 and 8 x + 7 y x y = 15 ⇒ 8 x x y + 7 y x y = 15 ⇒ 8 y + 7 x = 15 Let 1 x = u , 1 y = v 7 v - 2 u = 5 . . . . . ( i ) 8 v + 7 u = 15 . . . . . . ii Multiply eq i with 7 and eq ii with 2 . 49 v - 14 u = 35 . . . . . . iii 16 v + 14 u = 30 . . . . . . ( iv ) Adding iii & iv 65 v = 65 ⇒ v = 1 And 1 y = v = 1 ⇒ y = 1 Putting the value of v in eq i 7 ( 1 ) - 2 u = 5 ⇒ u = 1 And 1 x = u = 1 ⇒ x = 1 ∴ ( x , y ) = ( 1 , 1 )

Solve the following simultaneous equations. 1 2 ( 3 x + 4 y ) + 1 5 ( 2 x - 3 y ) = 1 4 ; 5 ( 3 x + 4 y ) - 2 ( 2 x - 3 y ) = - 3 2

Given: 1 2 ( 3 x + 4 y ) + 1 5 ( 2 x - 3 y ) = 1 4 ; 5 ( 3 x + 4 y ) - 2 ( 2 x - 3 y ) = - 3 2 Let 1 3 x + 4 y = u & 1 2 x - 3 y = v 1 2 u + 1 5 v = 1 4 ⇒ 5 u + 2 v 10 = 1 4 ⇒ 10 u + 4 v = 5 . . . . . . i 5 u - 2 v = - 3 2 ⇒ 10 u - 4 v = - 3 . . . . . . ( ii ) On adding i & ii 20 u = 2 ⇒ u = 1 10 Putting the value of u in ii 10 × 1 10 - 4 v = - 3 ⇒ 1 + 3 = 4 v ⇒ v = 1 Since, 1 3 x + 4 y = u = 1 10 ⇒ 3 x + 4 y = 10 . . . . . . iii And 1 2 x - 3 y = v = 1 ⇒ 2 x - 3 y = 1 . . . . . . ( iv ) Multiply eq iii with 2 and iv with 3 6 x + 8 y = 20 . . . . . . ( v ) 6 x - 9 y = 3 . . . . . . vi Now eq v - vi 17 y = 17 ⇒ y = 1 Putting the value of y in vi 6 x - 9 = 3 ⇒ 6 x = 12 ⇒ x = 2 ∴ ( x , y ) = ( 2 , 1 )

E M B I B E

Algebra Standard X>Chapter 1 - Linear Equations in Two Variables>Problem Set>Q 6

HARD

10th Maharashtra Board

IMPORTANT

Earn 100

Solve the following simultaneous equations.

$\frac{7}{2 x + 1} + \frac{13}{y + 2} = 27; \frac{13}{2 x + 1} + \frac{7}{y + 2} = 33$

Important Questions on Linear Equations in Two Variables

MEDIUM

10th Maharashtra Board

IMPORTANT

Algebra Standard X>Chapter 1 - Linear Equations in Two Variables>Problem Set>Q 6

Solve the following simultaneous equations.

$\frac{148}{x} + \frac{231}{y} = \frac{527}{x y}; \frac{231}{x} + \frac{148}{y} = \frac{610}{x y}$

HARD

10th Maharashtra Board

IMPORTANT

Algebra Standard X>Chapter 1 - Linear Equations in Two Variables>Problem Set>Q 6

Solve the following simultaneous equations.

$\frac{7 x - 2 y}{x y} = 5; \frac{8 x + 7 y}{x y} = 15$

HARD

10th Maharashtra Board

IMPORTANT

Algebra Standard X>Chapter 1 - Linear Equations in Two Variables>Problem Set>Q 6

Solve the following simultaneous equations.

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

HARD

10th Maharashtra Board

IMPORTANT

Algebra Standard X>Chapter 1 - Linear Equations in Two Variables>Problem Set>Q 7

Solve the following word problems.

A two-digit number and the number with digits interchanged add up to $143$ . In the given number the digit in the unit's place is $3$ more than the digit in the ten's place. Find the original number.

HARD

10th Maharashtra Board

IMPORTANT

Algebra Standard X>Chapter 1 - Linear Equations in Two Variables>Problem Set>Q 7

Kantabai bought

1 \frac{1}{2} kg

tea and

5 kg

sugar from a shop. She paid

₹ 50

as return fare for rickshaw. Total expense was

₹ 700

. Then she realised that by ordering online the goods can be bought with free home delivery at the same price. So next month she placed the order online for

2 kg

tea and

7 kg

sugar. She paid

₹ 880

for that. Find the rate of sugar and tea per

kg

.

MEDIUM

10th Maharashtra Board

IMPORTANT

Algebra Standard X>Chapter 1 - Linear Equations in Two Variables>Problem Set>Q 7

Anushka got $₹ 2500$ from Anand in the denomination of $₹ 100$ and $₹ 50$ . If Anand would have given her the amount by interchanging the number of notes, Anushka would have received $₹ 500$ less than the previous amount. Find the number of notes that Anushka had.

HARD

10th Maharashtra Board

IMPORTANT

Algebra Standard X>Chapter 1 - Linear Equations in Two Variables>Problem Set>Q 7

Sum of the present ages of Manish and Savita is