4x-7-(x+4)=3x+4-(2x-1)

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Solution

I guess the question is to solve the equation

4x - 7 - ( x + 4 ) = 3x + 4 - ( 2x - 1 )

we know that,
- ( a + b ) = -a - b
so ,
4x - 7 -x - 4 = 3x + 4 - ( 2x - 1 )

we have,
-( a -b ) = -a + b
so,
4x - 7 - x - 4 = 3x + 4 -2x + 1
4x - x -7 -4 = 3x -2x + 4 + 1

3x - 11 = x + 5

taking all the x terms on LHS and all the constants on RHS,
3x - x = 5 + 11
2x = 16
x = 8

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Similar questions

Which of these statements hold true?

i.) $4 (x - 5) = 4 x - 5$

ii.) $x (3 x + 2) = 3 x^{2} + 2$

iii.) $2 x^{2} + 4 (2 x) + 7 = 2 x^{2} + 8 x + 7$ [4 MARKS]

Q. What must be subtracted from

x^{4} + 4 x^{2} - 3 x + 7

to get

3 x^{3} - x^{2} + 2 x + 1

Answer:

x^{4} - 3 x^{3} + 5 x^{2} - 5 x + 6

Q. Divide

4 x^{4} - 3 x^{3} - 2 x^{2} + 6 x - 7

x - 2

Solve :

$(i) \frac{1}{3} x - 6 = \frac{5}{2} (i i) \frac{2 x}{3} - \frac{3 x}{8} = \frac{7}{12} (i i i) (x + 2) (x + 3) + (x - 3) (x - 2) - 2 x (x + 1) = 0 (i v) \frac{1}{10} - \frac{7}{x} = 35 (v) 13 (x - 4) - 3 (x - 9) - 4 (x + 4) = 0 (v i) x + 7 - \frac{8 x}{3} = \frac{17 x}{6} - \frac{5 x}{8} (v i i) \frac{3 x - 2}{4} - \frac{2 x + 3}{3} = \frac{2}{3} - x (v i i i) \frac{x + 2}{6} - (\frac{11 - x}{3} - \frac{1}{4}) = \frac{3 x - 4}{12} (i x) \frac{2}{5 x} - \frac{5}{3 x} = \frac{1}{15} (x) \frac{x + 2}{3} - \frac{x + 1}{5} = \frac{x - 3}{4} - 1 (x i) \frac{3 x - 2}{3} + \frac{2 x + 3}{2} = x + \frac{7}{6} (x i i) x - \frac{x - 1}{2} = 1 - \frac{x - 2}{3} (x i i i) \frac{9 x + 7}{2} - (x - \frac{x - 2}{7}) = 36 (x i v) \frac{6 x + 1}{2} + 1 = \frac{7 x - 3}{3}$

Q. Find the value of

x

x - [2 x - \frac{3 x - 4}{7}] = \frac{4 x - 27}{3} - 3

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