Question

solve the following pairs of linear equations by the method of substitution, 7x-15y-20=0 and x+2y= 3​

Answer 1

Answer:

x = 85 / 29

y = 3 - 2y

Step-by-step explanation:

Answer is in the attachment with complete steps.

Answer 2

Given :

• 7x - 15y - 20 = 0
• x + 2y = 3

To Find :

• Value of x by substitution method.

Solution :

We have :

• 7x - 15y - 20 = 0 - (1)
• x + 2y = 3 - (2)

From equation (1), we have :

$\sf : \implies 7x - 15y - 20 = 0$

$\sf : \implies 7x - 15y = 20$

$\sf : \implies 7x = 20 + 15y$

$\sf : \implies x = \dfrac{20 + 15y}{7}$

Substitute this value in equation (2) :

$\sf : \implies \dfrac{20 + 15y}{7} + 2y = 3$

$\sf : \implies \dfrac{20 + 15y}{7} + \dfrac{2y \times 7}{7} = 3$

$\sf : \implies \dfrac{20 + 15y}{7} + \dfrac{14y}{7} = 3$

$\sf : \implies \dfrac{20 + 15y + 14y}{7} = 3$

$\sf : \implies \dfrac{20 + 29y}{7} = 3$

$\sf : \implies 20 + 29y = 3 \times 7$

$\sf : \implies 20 + 29y = 21$

$\sf : \implies 29y = 21 - 20$

$\sf : \implies 29y = 1$

$\sf : \implies y = \dfrac{1}{29}$

$\Large \underline{\boxed{\bf{y = \dfrac{1}{29}}}}$

Now, by filling the value of y in equation (1), we get :

$\sf : \implies 7x - 15\Bigg(\dfrac{1}{29}\Bigg) - 20 = 0$

$\sf : \implies 7x - \dfrac{15}{29} - 20 = 0$

$\sf : \implies \dfrac{7x \times 29}{29} - \dfrac{15}{29} - \dfrac{20\times 29}{29} = 0$

$\sf : \implies \dfrac{203x}{29} - \dfrac{15}{29} - \dfrac{580}{29} = 0$

$\sf : \implies \dfrac{203x - 15 - 580}{29} = 0$

$\sf : \implies \dfrac{203x - 595}{29} = 0$

$\sf : \implies 203x - 595 = 0\times 29$

$\sf : \implies 203x - 595 = 0$

$\sf : \implies 203x = 595$

$\sf : \implies x = \cancel{\dfrac{595}{203}}$

$\sf : \implies x = \dfrac{85}{29}$

$\Large \underline{\boxed{\bf{x = \dfrac{85}{29}}}}$

Hence, value of :

• $\Large \underline{\boxed{\bf{x = \dfrac{85}{29}}}}$
• $\Large \underline{\boxed{\bf{y = \dfrac{1}{29}}}}$