hi everyone in this question we have to solve the pair of equation 2 X + 3 Y is equal to 13 and S is 5 by 8 minus 4 Y is equal to minus to get look at the solution part assume 1 by X U1 bi vi 8b question becomes let anatomy question please come to you that CV is equal to 13 and 5 you my hobby is equal to minus 2 X 1 into 58222 equation 1 and 2 were hired the question becomes a new love 15 V is equal to 55 and second become a new Show -8 V is equal to minus 4 3 this is for late payment 3 -4 just become can you get cancelled out here it is 23 we recall 269 the value of V comes out to be 2016 123 and putting the value of V and equation 1 wicket 2 + 3 into 3 equal to 13 the value of you comes out to be to know we have assumed 1 by X equal to 1 by x is equal to you used to value of x finally comes out to be one by two and one bye-bye has been won by Y is equal to be the value of Y is nothing but one by two are the final and access 1 by 2 and minus 1 by 3 thank you bhai Solve the following simultaneous equations. `2/x + 3/y = 13` ; `5/x - 4/y = -2` Concept: Algebraic Methods of Solving a Pair of Linear Equations - Elimination Method Solution: The standard form of a linear equation is ax + by + c = 0. To reduce a pair of equations to the standard form, we will use substitution. The equation 2/x + 3/ y = 13 can be expressed 2(1 /x) + 3(1/ y) = 13 be equation(1) The equation 5/x - 4/y = -2 can be expressed 5(1/x) - 4(1/ y) = - 2 be equation (2) Let 1/ x be a and 1/y be b. ⇒ 2a + 3b = 13 be equation (3) ⇒ 5a - 4b = - 2 be equation (4). Let us use the elimination method to find the values of a and b. Multiply equation (3) by 5 and equation (4) by 2 ⇒ 10a + 15b = 65 be equation (5) ⇒ 10a - 8b = - 4 be equation (6) Subtract equation (6) from equation (5) ⇒ (10a + 15b = 65 ) - (10a - 8b = - 4) ⇒ 23b = 69 ⇒ b = 69/ 23 ⇒ b = 3 Substitute the value of b = 3 in equation (3) ⇒ 2a + 3(3) = 13 ⇒ 2a + 13 - 9 ⇒ 2a = 4 ⇒ a = 2 ∵ a = 2; b = 3 ∴ the value of x = ½; y = ⅓ ☛ Check: NCERT Solutions for Class 10 Maths Chapter 3 Solve the pair of equations: 2/x + 3/y = 13, 5/x - 4/y = -2Summary: The values of x and y for the pair of linear equations is ½ and ⅓ which satisfies the equation ☛ Related Questions:
Example 17 - Chapter 3 Class 10 Pair of Linear Equations in Two Variables (Term 1)Last updated at Dec. 18, 2020 by
 This video is only available for Teachoo black users TranscriptExample 17 Solve the pair of equations: 2/𝑥+ 3/𝑦=13 5/𝑥−4/𝑦=−2 2/𝑥+ 3/𝑦=13 5/𝑥−4/𝑦=−2 So, our equations become 2u + 3v = 13 5u – 4v = –2 Hence, our equations are 2u + 3v = 13 …(3) 5u – 4v = – 2 …(4) From (3) 2u + 3v = 13 2u = 13 – 3V u = (13 − 3𝑣)/2 Putting value of u (4) 5u – 4v = - 2 5((13 − 3𝑣)/2)−4𝑣=−2 Multiplying 2 both sides 2 × 5((13 − 3𝑣)/2)−"2 ×" 4𝑣="2 ×"−2 5(13 – 3v) – 8v = –4 65 – 15v – 8v = –4 – 15v – 8v = – 4 – 65 – 23v = – 69 v = (−69)/(−23) v = 3 Putting v = 3 in (3) 2u + 3v = 13 2u + 3(3) = 13 2u + 9 = 13 2u = 13 – 9 2u = 4 u = 4/2 u = 2 Hence, u = 2, v = 3 is the solution But we have to find x & y u = 𝟏/𝒙 2 = 1/𝑥 x = 𝟏/𝟐 v = 𝟏/𝒚 3 = 1/𝑦 y = 𝟏/𝟑 Hence, x = 1/2 , y = 1/3 is the solution of the given equation |
