0.05x + 0.2y = 0.07

Let us consider x be 1 and y be 1/10

0.05(1)+0.2(1/10)=0.07

0.05+0.02=0.07

0.07=0.07

LHS=RHS

0.3x-0.1y=0.03

let x be 1/10 and y be 0

0.3(1/10)-0.1(0)=0.03

0.03-0=0.03

0.03=0.03

LHS=RHS.

Solving the Equations:

To solve the given equations, we need to find the values of x and y that satisfy both of them simultaneously. We can use various methods to solve the equations, such as substitution or elimination method.

Using Substitution Method:

In this method, we isolate one of the variables in one equation and substitute its value in the other equation to get a linear equation in one variable.

Let's isolate y in the second equation:

0.3x - 0.1y = 0.03

-0.1y = -0.3x + 0.03

y = 3x - 0.3

Now, substitute this value of y in the first equation:

0.05x + 0.2(3x - 0.3) = 0.07

0.05x + 0.6x - 0.06 = 0.07

0.65x = 0.13

x = 0.2

Substitute this value of x in the equation we got for y:

y = 3(0.2) - 0.3

y = 0.3

Hence, the solution of the given system of equations is (0.2, 0.3).

Checking the Solution:

We can check whether the obtained values of x and y satisfy both the equations by substituting them in the original equations.

0.05(0.2) + 0.2 = 0.07 (True)

0.3(0.2) - 0.1(0.3) = 0.03 (True)

Hence, the solution (0.2, 0.3) is correct.

Summary:

- The given system of equations is 0.05x + 0.2 = 0.07 and 0.3x - 0.1y = 0.03.
- We can solve the equations using various methods, such as substitution or elimination method.
- Using substitution method, we obtained the solution (0.2, 0.3).
- We checked the solution by substituting the obtained values in the original equations.