To solve a simultaneous equation with 2 variables, one easier way is to multiply any one of the equation by a constant, such that the coefficient of the same variable in the 2 equations becomes the same.
In this example, multiply the second equation by $2$. We get the system
$$a+b/3=1$$
$$a+b/2=3/5$$
Then subtract one equation from another. As one may observe, one of the variable vanishes. This is called elimination.
Then one should obtain an equation with one variable left. To be more specific here, an equation in terms of only $b$ is produced. Solving the equation, one can obtain the value of the variable.
By back substitution, that is, to plug in the solved value of the variable (Here, $b$ is the solved one.) into any of the equations, the other unknown can finally be solved.
A more advanced yet general and equivalent method is to reduce the coefficent matrix of the system into row-echelon form.
To solve a simultaneous equation with 2 variables, one easier way is to multiply any one of the equation by a constant, such that the coefficient of the same variable in the 2 equations becomes the same.
In this example, multiply the second equation by $2$. We get the system $$a+b/3=1$$ $$a+b/2=3/5$$
Then subtract one equation from another. As one may observe, one of the variable vanishes. This is called elimination.
Then one should obtain an equation with one variable left. To be more specific here, an equation in terms of only $b$ is produced. Solving the equation, one can obtain the value of the variable.
By back substitution, that is, to plug in the solved value of the variable (Here, $b$ is the solved one.) into any of the equations, the other unknown can finally be solved.
A more advanced yet general and equivalent method is to reduce the coefficent matrix of the system into row-echelon form.