Q: What branch of maths involves solving on x?

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In algebra whereas x is usually the unknown variable to be solved.

It is usually algebra whereas x is an unknown variable.

Solving a one variable linear equation involves getting the variable on one side of the equals sign by itself. To do this one uses the properties of numbers.

Solving 11x-28=82x for x involves the following: -28=82x-11x -28=71x -28/71=x x=.3943 (truncated to 4 decimal places)

It depends on what you are solving for. If you are solving for 'y', then the answer would be: y=4x+2 if you're solving for 'x': x=1/4y-1

Related questions

In algebra whereas x is usually the unknown variable to be solved.

It is usually algebra whereas x is an unknown variable.

Solving a one variable linear equation involves getting the variable on one side of the equals sign by itself. To do this one uses the properties of numbers.

Solving 11x-28=82x for x involves the following: -28=82x-11x -28=71x -28/71=x x=.3943 (truncated to 4 decimal places)

x to x +25

If you multiply each term of the first type of equation by a common multiple of all the denominators then you will have an equation of the second type.For example, if you have 2/3*y = 4/5*x + 7/9 then multiplying by the LCM of 3, 5, 9) = 45, gives30*y = 39*x + 35: all integers!

If you are solving for y, it is fine. If you are solving for x, divide both sides by x and the equation should be x = y/x

It depends on what you are solving for. If you are solving for 'y', then the answer would be: y=4x+2 if you're solving for 'x': x=1/4y-1

250x + 250 * x.

The cast of Solving for X - 2009 includes: Bill Nye as himself

Either use trial and error, or the quadratic formula, solving the following for x: x(x+1)=182Either use trial and error, or the quadratic formula, solving the following for x: x(x+1)=182Either use trial and error, or the quadratic formula, solving the following for x: x(x+1)=182Either use trial and error, or the quadratic formula, solving the following for x: x(x+1)=182

An unknown number.