### Related

##### breakup-solve

eq1:V = F * ((C*(1 - ...

breakup:falseres:solv...

Calculate

##### breakup-false-programmode-solve-true

eq1:V = F * ((C*(1 - ...

programmode: false;

breakup: true;

Calculate

##### breakup-false-solve

eq1:V = F * ((C*(1 - ...

breakup:false;

res:solve(eq1, Y);

Calculate

##### breakup-false-programmode-solve-true

programmode: false;

breakup: true;

solve (x^3 + x^2 - 1);

Calculate

##### breakup-false-multiplicities-programmode-solve

eq1:V = F * ((C*(1 - ...

programmode: false;

breakup: false;

Calculate

##### breakup-false-programmode-solve

eq1:V = F * ((C*(1 - ...

programmode: false;

breakup: false;

Calculate

##### breakup-false-programmode-solve-true

programmode: false;

breakup: true;

solve (x^3 + x^2 - 1);

Calculate

##### breakup-false-programmode-solve-true

programmode: false;

breakup: true;

solve (x^3 + x^2 - 1);

Calculate

##### breakup-false-programmode-solve-true

eq1:V = F * ((C*(1 - ...

programmode: false;

breakup: true;

Calculate

? breakup;

Calculate

### breakup

Run Example
```(%i1)eq1:V = F * ((C*(1 - (1/(1 + Y))^(M*H)) / H*Y) + (1 / (1 + Y))^(M*H));
1
C Y (1 - ----------)
H M
1                 (Y + 1)
(%o1)              V = F (---------- + --------------------)
H M            H
(Y + 1)
(%i2) breakup:falseres:solve(eq1, Y);
H M
H V (Y + 1)    - F H
(%o2)                     [Y = --------------------]
H M
C F (Y + 1)    - C F
(%i3) ```
Run Example
```? breakup;

-- Option variable: breakup
Default value: `true'

When `breakup' is `true', `solve' expresses solutions of cubic and
quartic equations in terms of common subexpressions, which are
assigned to intermediate expression labels (`%t1', `%t2', etc.).
Otherwise, common subexpressions are not identified.

`breakup: true' has an effect only when `programmode' is `false'.

Examples:

(%i1) programmode: false\$
(%i2) breakup: true\$
(%i3) solve (x^3 + x^2 - 1);

sqrt(23)    25 1/3
(%t3)                  (--------- + --)
6 sqrt(3)   54
Solution:

sqrt(3) %i   1
---------- - -
sqrt(3) %i   1            2        2   1
(%t4)    x = (- ---------- - -) %t3 + -------------- - -
2        2            9 %t3        3

sqrt(3) %i   1
- ---------- - -
sqrt(3) %i   1              2        2   1
(%t5)    x = (---------- - -) %t3 + ---------------- - -
2        2             9 %t3         3

1     1
(%t6)                  x = %t3 + ----- - -
9 %t3   3
(%o6)                    [%t4, %t5, %t6]
(%i6) breakup: false\$
(%i7) solve (x^3 + x^2 - 1);
Solution:

sqrt(3) %i   1
---------- - -
2        2        sqrt(23)    25 1/3
(%t7) x = --------------------- + (--------- + --)
sqrt(23)    25 1/3    6 sqrt(3)   54
9 (--------- + --)
6 sqrt(3)   54

sqrt(3) %i   1    1
(- ---------- - -) - -
2        2    3

sqrt(23)    25 1/3  sqrt(3) %i   1
(%t8) x = (--------- + --)    (---------- - -)
6 sqrt(3)   54          2        2

sqrt(3) %i   1
- ---------- - -
2        2      1
+ --------------------- - -
sqrt(23)    25 1/3   3
9 (--------- + --)
6 sqrt(3)   54

sqrt(23)    25 1/3             1             1
(%t9)  x = (--------- + --)    + --------------------- - -
6 sqrt(3)   54          sqrt(23)    25 1/3   3
9 (--------- + --)
6 sqrt(3)   54
(%o9)                    [%t7, %t8, %t9]

(%o1)                                true
(%i2) ```
Run Example
```breakup: false;
(%o1)                                false
(%i2) eq1:1/(2*%pi*sigma^2) * exp(-(x-mu_x)^2+(y-mu_y)^2)/(2*sigma^2);
2             2
(y - mu_y)  - (x - mu_x)
%e
(%o2)                     ---------------------------
4
4 %pi sigma
(%i3) eq2:1/(2*%pi*sigma1^2) * exp(-(x-mu_x1)^2+(y-mu_y1)^2)/(2*sigma1^2);
2              2
(y - mu_y1)  - (x - mu_x1)
%e
(%o3)                    -----------------------------
4
4 %pi sigma1
(%i4) algsys([eq1,eq2],[x,y]);
(%o4)                                 []
(%i5) ```

Help for Breakup