Question

17. Find the values of x and y in the given equations: (5 marks)
(a) 3x + 5y = 21 and 2x + 3y = 13 (Use cancellation method)
(b) 5x – 2y = 17 and 3x + y = 8 (Use substitution method)​

Answer 1

Answer:

(a) x=2,y=3

(b)x=3,y=-1

Step-by-step explanation:

(a)

3x+5y=21   -(1)

2x+3y=13  -(2)

multiply 1st eqn by 2,

6x+10y=42

multiply 2nd eqn by 3

6x+9y=39

subtract

6x+10y-6x-9y=42-39

y=3

substitute in (1)

3x+15=21

3x=6, x=2

(b)

5x-2y=17  -(1)

3x+y=8   -(2)

from (2), y=8-3x

substitute in (1)

5x-2(8-3x)=17

5x-16+6x=17

11x=33, x=3

put value of x in (2)

9+y=8

y=-1

Answer 2

★ The solution of the first pair of equations is x = 2 and y = 3, whereas the solution of the second pair of equations is x = 3 and y = –1.

Step-by-step explanation

Analysis -

‎ ‎ ‎ ‎ ‎ ‎ ‎In the question, it has been given that we have to find the value of the variables x and y in two simultaneous equations. And a condition has also been mentioned that the first set of equations must be solved by cancellation method and second set by substitution method.

Solution -

‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎Generally, when two equations, each in two variables are given, they can be solved in four ways.

1. Elimination by cancellation.
2. Elimination by substitution.
3. Adding the two equations and subtracting one equation from the other.
4. Graphical method.

As per the condition, we will be using the first two methods for the given set of equations, respectively. Let us start solving!

‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎Elimination by Cancellation

‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎(3x + 5y = 21) and (2x + 3y = 13)

Using this method, the two equations are reduced to a single variable equation by eliminating one of the variables.

Step 1:

Consider (3x + 5 = 21) as the first equation (eq. 1) and (2x + 3y = 12) as the second equation (eq. 2).

Step 2:

Here, let us we eliminate the y term and in order to eliminate the y term, multiply the first equation with the coefficient of y in the second equation and multiply the second equation with the coefficient of y in the first equation, so that the coefficients of y terms in the both the equations becomes equal.

$\dashrightarrow{ \sf \pmb{3(3x + 5y = 21)}} \\ \\ \Rightarrow{ \sf{9x + 15y = 65 \: \textsf{ - - - } \: (eq. \: 3)}} \\ \\ \\ \dashrightarrow{ \sf \pmb{5(2x + 3y = 13)}} \\ \\ \Rightarrow{ \sf{10x + 15y = 65 \: \textsf{ - - - } \: (eq. \: 4)}}$

Step 3:

Subtract (eq. 3) from (eq. 4).

$\dashrightarrow { \sf{(10x + 15y) - (9x + 15y) = 65 - 63}} \\ \\ \Rightarrow \underline{ \boxed{ \sf \pmb{x = \frak{2}}}}$

Step 4:

Substitute the value of x in (eq. 1) or (eq. 2) to find the value of y.

Here, substituting the value of x in the first equation we have,

$\dashrightarrow{ \sf{3x + 5y = 21}} \\ \\ \dashrightarrow{ \sf{3(2) + 5y = 21}} \\ \\ \dashrightarrow{ \sf{6 + 5y = 21}} \\ \\ \dashrightarrow{ \sf{5y = 21 - 6}} \\ \\ \dashrightarrow{ \sf{5y = 15}} \\ \\ \dashrightarrow{ \sf{y = \dfrac{ \cancel{15}}{ \cancel5} }} \\ \\ \dashrightarrow \underline{ \boxed{ \sf{ \pmb{ y = \frak 3}}}}$

Now we have both the values of x and y.

Therefore, the solution of the given set of equations is (x = 2) and (y = 3).

‎

‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎Elimination by Substitution

‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎(5x – 2y = 17) and (3x + y = 8)

Using this method, the two equations are reduced to a single variable equation by substituting the value of one variable obtained from one equation in the other equation.

Step 1:

Consider (5x – 2y = 17) as the first equation (eq. 1) and (3x + y = 8) as the second equation (eq. 2).

Step 2:

Using the first equation, find x in terms of y.

$\dashrightarrow{ \sf{5x - 2y = 17}} \\ \\ \dashrightarrow{ \sf{5x = 17 + 2y}} \\ \\ \dashrightarrow{ \sf{ x = \dfrac{17 + 2y}{5} }}$

Step 3:

Substitute the obtained value of x in the (eq. 2) to find the value of y.

$\dashrightarrow{ \sf{3x + y = 8}} \\ \\ \dashrightarrow{ \sf{3 \bigg( \dfrac{17 + 2y}{5} \bigg) + y =8 }}$

Step 4:

Simplify the equation in terms of y and find the value of y.

$\dashrightarrow{ \sf{3 \bigg( \dfrac{17 + 2y}{5} \bigg) + y =8 }} \\ \\ \dashrightarrow{ \sf{3(17 + 2y) + 5y = 8(5)}} \\ \\ \dashrightarrow{ \sf{51 +6 y + 5y = 40}} \\ \\ \dashrightarrow{ \sf{51 + 11y = 40}} \\ \\ \dashrightarrow{ \sf{11y = 40 - 51}} \\ \\ \dashrightarrow{ \sf{11y = - 11}} \\ \\ \dashrightarrow{ \sf{y = \cancel \dfrac{ - 11}{11} }} \\ \\ \dashrightarrow \underline{ \boxed{ \sf \pmb{y = \frak{ - 1}}}}$

Step 5:

Substituting the value of y obtained in step 4 in (eq. 1 or 2), we get,

$\dashrightarrow{ \sf{5x - 2y = 17}} \\ \\ \dashrightarrow{ \sf{5x - 2( - 1) = 17}} \\ \\ \dashrightarrow{ \sf{5x - ( - 2) = 17}} \\ \\ \dashrightarrow{ \sf{5x + 2 = 17}} \\ \\ \dashrightarrow{ \sf{5x = 17 - 2}} \\ \\\dashrightarrow{ \sf{5x = 15}} \\ \\ \dashrightarrow{ \sf{x = \dfrac{ \cancel{15}}{ \cancel5} }} \\ \\ \dashrightarrow \underline{ \boxed{ \sf \pmb{x = \frak{3}}}}$

Therefore, the solution of the given pair of equations is (x = 3) and (y = –1).

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