Question

subtract the sum of 3 x minus 4 y + z and 1 - 2 x + y from the sum of 3 y + z + 6 and x - 2y + 3​

Answer 1

$\huge\tt{\bold{\underline{\underline{Question᎓}}}}$

subtract the sum of 3 x minus 4 y + z and 1 - 2 x + y from the sum of 3 y + z + 6 and x - 2y + 3

$\huge\tt{\bold{\underline{\underline{Answer᎓}}}}$

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$= > 3x - 4y + z + 1 - 2x + y$

$= 3x - 2x - 4y + y + z + 1$

$\bold{= x - 3y + z + 1.......(1)}$

$= > 3y + z + 6 + x - 2y + 3$

$= x + 3y - 2y + z + 9$

$\bold{= x + y + z + 9.......(2)}$

Now,subtracting equation (1) from equation (2)

$= > x + y + z + 9 - (x - 3y + z + 1)$

$= x + y + z + 9 - x + 3y - z - 1$

=>Here,when we open bracket -x and -z will come which will being cancelled by existing +ve x and z.

$\bold{= > 4y + 8 = }{\bold{\red{4(y + 2)}}}$

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