`solve`

FunctionIf `solve`

returns solutions that look complicated,
or if `solve`

cannot handle an input, there are
many options. These options simplify the solution space for `solve`

.
These options also help `solve`

when the input
is complicated, and might allow `solve`

to return
a solution where it was previously stuck.

Solve the equation `x^5 - 1 == 0`

. This equation
has five solutions.

syms x solve(x^5 - 1 == 0, x)

ans = 1 - (2^(1/2)*(5 - 5^(1/2))^(1/2)*1i)/4 - 5^(1/2)/4 - 1/4 (2^(1/2)*(5 - 5^(1/2))^(1/2)*1i)/4 - 5^(1/2)/4 - 1/4 5^(1/2)/4 - (2^(1/2)*(5^(1/2) + 5)^(1/2)*1i)/4 - 1/4 5^(1/2)/4 + (2^(1/2)*(5^(1/2) + 5)^(1/2)*1i)/4 - 1/4

If you only need real solutions, specify the `Real`

option
as `true`

. The `solve`

function
returns the one real solution.

solve(x^5 - 1, x, 'Real', true)

ans = 1

Solve the following equation. The `solve`

function
returns a complicated solution.

syms x solve(x^(5/2) + 1/x^(5/2) == 1, x)

ans = 1/(1/2 - (3^(1/2)*1i)/2)^(2/5) 1/((3^(1/2)*1i)/2 + 1/2)^(2/5) -(5^(1/2)/4 - (2^(1/2)*(5 - 5^(1/2))^(1/2)*1i)/4 + 1/4)/(1/2 - (3^(1/2)*1i)/2)^(2/5) -((2^(1/2)*(5 - 5^(1/2))^(1/2)*1i)/4 + 5^(1/2)/4 + 1/4)/(1/2 - (3^(1/2)*1i)/2)^(2/5) -(5^(1/2)/4 - (2^(1/2)*(5 - 5^(1/2))^(1/2)*1i)/4 + 1/4)/(1/2 + (3^(1/2)*1i)/2)^(2/5) -((2^(1/2)*(5 - 5^(1/2))^(1/2)*1i)/4 + 5^(1/2)/4 + 1/4)/(1/2 + (3^(1/2)*1i)/2)^(2/5)

To apply simplification rules when solving equations, specify
the `IgnoreAnalyticConstraints`

option as `true`

.
The applied simplification rules are not generally correct mathematically
but might produce useful solutions, especially in physics and engineering.
With this option, the solver does not guarantee the correctness and
completeness of the result.

solve(x^(5/2) + 1/x^(5/2) == 1, x, 'IgnoreAnalyticConstraints', true)

ans = 1/(1/2 - (3^(1/2)*1i)/2)^(2/5) 1/((3^(1/2)*1i)/2 + 1/2)^(2/5)

This solution is simpler and more usable.

For solutions to specific cases, set assumptions to return appropriate
solutions. Solve the following equation. The `solve`

function
returns seven solutions.

syms x solve(x^7 + 2*x^6 - 59*x^5 - 106*x^4 + 478*x^3 + 284*x^2 - 1400*x + 800, x)

ans = 1 - 5^(1/2) - 1 - 17^(1/2)/2 - 1/2 17^(1/2)/2 - 1/2 -5*2^(1/2) 5*2^(1/2) 5^(1/2) - 1

Assume `x`

is a positive number and solve the
equation again. The `solve`

function only returns
the four positive solutions.

assume(x > 0) solve(x^7 + 2*x^6 - 59*x^5 - 106*x^4 + 478*x^3 + 284*x^2 - 1400*x + 800, x)

ans = 1 17^(1/2)/2 - 1/2 5*2^(1/2) 5^(1/2) - 1

Place the additional assumption that `x`

is
an integer using `in(x,'integer')`

. Place additional
assumptions on variables using `assumeAlso`

.

assumeAlso(in(x,'integer')) solve(x^7 + 2*x^6 - 59*x^5 - 106*x^4 + 478*x^3 + 284*x^2 - 1400*x + 800, x)

ans = 1

`solve`

returns the only positive, integer
solution to `x`

.

Clear the assumptions on `x`

for further computations by recreating it using
`syms`

.

syms x

Alternatively, to make several assumptions, use the `&`

operator.
Make the following assumptions, and solve the following equations.

syms a b c f g h y assume(f == c & a == h & a~= 0) S = solve([a*x + b*y == c, h*x - g*y == f], [x, y], 'ReturnConditions', true); S.x S.y S.conditions

ans = f/h ans = 0 ans = b + g ~= 0

Under the specified assumptions, the solution is ```
x
= f/h
```

and `y = 0`

under the condition ```
b
+ g ~= 0
```

.

Clear the assumptions on the variables for further computations by recreating them using
`syms`

.

syms a c f h

The `solve`

function does not call simplification
functions for the final results. To simplify the solutions, call `simplify`

.

Solve the following equation. Convert the numbers to symbolic
numbers using `sym`

to return a symbolic result.

syms x S = solve((sin(x) - 2*cos(x))/(sin(x) + 2*cos(x)) == 1/2, x)

S = -log(-(- 140/37 + 48i/37)^(1/2)/2)*1i -log((- 140/37 + 48i/37)^(1/2)/2)*1i

Call `simplify`

to simplify solution `S`

.

simplify(S)

ans = -log(37^(1/2)*(- 1/37 - 6i/37))*1i log(2)*1i - (log(- 140/37 + 48i/37)*1i)/2

Call `simplify`

with more steps to simplify
the result even further.

simplify(S, 'Steps', 50)

ans = atan(6) - pi atan(6)

To represent a number exactly, use

`sym`

to convert the number to a floating-point object. For example, use`sym(13)/5`

instead of`13/5`

. This represents`13/5`

exactly instead of converting`13/5`

to a floating-point number. For a large number, place the number in quotes. Compare`sym(13)/5`

,`sym(133333333333333333333)/5`

, and`sym('133333333333333333333')/5`

.sym(13)/5 sym(133333333333333333333)/5 sym('133333333333333333333')/5

ans = 13/5 ans = 133333333333333327872/5 ans = 133333333333333333333/5

Placing the number in quotes and using

`sym`

provides the highest accuracy.If possible, simplify the system of equations manually before using

`solve`

. Try to reduce the number of equations, parameters, and variables.