Question

3x+2y=6
4x+7y=12
Solve the equation​

Answer 1

Answer:

Reduce the following equations into intercept form and find their intercepts on

the axes.

(i)3x+2y−12=0

(ii)4x−3y=6

(iii)3y+2=0

Answer 2

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Given

• System of equations
• 3x+2y=6----------(1)
• 4x+7y=12---------(2)

To find

• value of x and y

Solution

Solving the above equation by elimination method

Multiplying the eq(1) by 4 and eq(2) by 4 we get

• $12x + 8y = 24 - - - - - -eq(3)$
• $12x + 21y = 36 - - - - - eq(4)$

Now subtracting eq(3) from eq(4) we get

$12x + 21y = 36 \\ 12x + 8y \: \: = 24 \\ - \: \: \: \: \: - \: \: \: \: \: \: \: \: \: - \\ 0 \: \: \: \: \: \: \: + 13y = 12$

Therefore

$y = \frac{12}{13} - - - - - - eq(5)$

Now putting y= 12/13 in eq(1)

$3x + 2( \frac{12}{13}) = 6$

$3x = 6 - \frac{24}{13}$

$x = \frac{54}{13} \div 3$

$= > x = \frac{18}{13}$

Hence

$x = \frac{18}{13} \: \: \: and \: \: y = \frac{12}{13}$

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