Question

Solve the system by the method of substitution. $$\left\\{\begin{array}{l}0.3 x-0.4 y-0.33=0 \\ 0.1 x+0.2 y-0.21=0\end{array}\right.$$

Short answer

Hence, the solution of the system of equations is \( x = 0.1 \) and \( y = -0.75 \)

Step-by-step solution

Isolate one variable from one of the equations

Let's simplify the first equation \(0.3x - 0.4y - 0.33=0\) by isolating variable 'x'. To do that, add \(0.4y\) and \(0.33\) to both sides of the equation, obtaining \(0.3x = 0.4y + 0.33\). Then dividing all terms by \(0.3\), we get \(x = \frac{4y+3.3}{3}\). This way, 'x' is expressed in terms of 'y'.

Substitute the isolated variable into the second equation

Substitute \(x = \frac{4y+3.3}{3}\) into the second equation: \(0.1x + 0.2y - 0.21 = 0\). This gives \(0.1(\frac{4y+3.3}{3}) + 0.2y - 0.21 = 0\). Multiply through by 10 to clear out the decimal, we get \((\frac{4y+3.3}{3}) + 2y - 2.1 = 0.\) Simplify the equation by making y the subject.

Solve the equation for y

Combine like terms and simplifying the resulting equation, we have \(\frac{4y+3.3}{3} + 2y = 2.1\). Subtract \(2y\) from both sides to get \(\frac{4y+3.3}{3} = 0.1\). Multiply through by 3 to clear the fraction: \(4y + 3.3 = 0.3\). Subtract \(3.3\) from both sides to get \(4y = -3\). Divide by 4 to get \(y=-0.75\)

Substitute y into the isolated variable equation

Substitute \(y = -0.75\) into the isolated variable equation \(x = \frac{4y+3.3}{3}\), we get \(x = \frac{4*(-0.75)+3.3}{3} \Rightarrow x = 0.1 \)

Key concepts

Solving systems of equations

Solving systems of equations means finding values for the variables that satisfy all the equations simultaneously. There are several methods to solve them, such as graphing, substitution, and elimination. In this exercise, the substitution method was used.

Here's how substitution works step by step:

• Choose one of the equations and solve for one variable in terms of the other.
• Substitute that expression into the other equation.
• Solve the resulting equation to find the value of the second variable.
• Use that value to find the value of the first variable.

Substitution is particularly useful when one of the equations is easy to solve for a single variable.
This method reduces the system to a single equation with one unknown variable, making it simpler to solve.

Linear equations

Linear equations are equations of the first degree, meaning they have no variables raised to a power higher than one. The general form of a linear equation is \( ax + by = c \), where \(a\), \(b\), and \(c\) are constants. In a two-variable system, these equations graph as straight lines, and the point of intersection represents the solution.

For example, in the system given, both equations are linear:

• First equation: \(0.3x - 0.4y - 0.33= 0 \)
• Second equation: \(0.1x + 0.2y - 0.21= 0\)

Each equation depicts a line, and solving the system will determine where these lines intersect. Linear systems with two variables generally have one solution (the lines intersect at one point), no solutions (the lines are parallel), or infinitely many solutions (the lines coincide).

Algebraic manipulation

Algebraic manipulation involves rearranging and simplifying equations to make them easier to work with. This process is crucial for solving systems of equations using the substitution method.

Here are some common algebraic manipulation steps:

• Moving terms from one side of an equation to another by adding or subtracting.
• Dividing or multiplying entire equations to simplify or clear fractions.
• Substituting values or expressions to reduce the number of variables in an equation.
  This helps in isolating one variable to express it in terms of another.

This exercise required manipulating the first equation to express \(x\) in terms of \(y\) and using that expression in the second equation to finally find the values of both variables. Remember, practicing algebraic manipulation will make solving any system of equations faster and more intuitive.