Gọi `m_(\text{dd HCl 10%}) = x (g)`, `m_(\text{dd HCl 25%}) = y (g)`

Ta có : `m_(\text{dd HCl 19%}) = m_(\text{dd HCl 10%}) + m_(\text{dd HCl 25%})`

`=> x + y = 500` `(1)`

`m_(\text{dd HCl 19%}) = 500g`

`=> m_(\text{HCl(dd HCl 19%)}) = 500 . 19% = 95g`

`m_(\text{dd HCl 10%}) = x (g)`

`=> m_(\text{HCl(dd HCl 10%)}) = x . 10% = 0,1x(g)`

`m_(\text{dd HCl 25%}) = y (g)`

`=> m_(\text{HCl(dd HCl 25%)}) = y . 25% = 0,25y(g)`

Có : `m_(\text{HCl(dd HCl 25%)}) + m_(\text{HCl(dd HCl 10%)}) = m_(\text{HCl(dd HCl 19%)})`

`=> 0,1x + 0,25y = 95` `(2)`

Giải PT `(1)` và `(2)` ta được : `x = 200, y = 300`

Vậy cô Linh cần `200g` dd `HCl` `10%` và `300g` dd `HCl` `25%`