For the following expressions,why we can get the right answer even we do addition /subtraction first ?
$3+4\times 11-5 = (3+4)\times(11-5) = 42$
$6+4\times7-4$
$5+2\times13-10$
$4+7\times16-6$
$3-2\times1+5$
For the following expressions,why we can get the right answer even we do addition /subtraction first ?
$3+4\times 11-5 = (3+4)\times(11-5) = 42$
$6+4\times7-4$
$5+2\times13-10$
$4+7\times16-6$
$3-2\times1+5$
Expressed algebraically, you have:
$$\begin{align} a+bc+d&=(a+b)(c+d)\\ a+bc+d&=ac+ad+bc+bd\\ a+d-ac-ad-bd&=0\\ a(1-c-d)+d(1-b)&=0\\ \end{align}$$
This is a single simultaneous equation in 4 unknowns, therefore it has an infinite number of solutions - you have just found some of them.
Rewriting to address OP comment
$$a=\frac{d(b-1)}{1-c-d}$$
Pick any integers $c$ and $d$. Choose integer $b$ such that $a$ is an integer.