Solving Linear Equations

Solve the equation$3x+2=0$.

Solution

$3x+2=0$

Each side of the equation is simplified, step 1 is achieved.

Step 2: Group like terms by subtracting 2 from each side of the equation

$3x+2-2=0-2$

$3x=-2$

Step 3: Divide each side by 3

$\frac{3x}{3}=\frac{-2}{3}$

$x=\frac{-2}{3}$

Step 4: Now we can evaluate this to see if this is true. The equation means that everything on the left side should be equal to what is on the right. Hence, everything on the left side of the equation should be equal to 0. We will substitute the solution into the equation now.

$$\begin{gathered} 3x+2=0 \\ 3\left(\frac{-2}{3}\right)+2=0 \end{gathered}$$

We will now divide 3 outside the bracket by the 3 as the denominator, and we will have 1 each.

$-2+2=0$

$0=0$

We see here that the solution we have is true.

Solve for x and y given the equations $2x+5y=20$ and $3x+5y=12$.

Solution

$\left\{\begin{array}{l}2x+5y=20\\ 3x+5y=12\end{array}\right.$

Let us make y the subject of the first equation by subtracting 2x from each side of the equation.

$2x-2x+5y=20-2x$

Now we will substitute this value of y into the second equation

$3x+5y=12$

We will now substitute this value for x into any of the equations to find y. We will use the first.

$2x+5y=20$

Add 16 to each side of the equation to make 5y stand alone on that side of the equation

$16-16+5y=20+16$

Divide through by 5 to find y

$\frac{5y}{5}=\frac{36}{5}$

$y=\frac{36}{5}$

Solve the equation

$\left\{\begin{array}{l}y-2x=2\\ -x=-y-1\end{array}\right.$

Solution

As mentioned earlier, we will want to plot both equations on the coordinate plane. We will start by finding the y-intercept and slope for each line. This means for each equation, we will rewrite it in the slope-intercept form. Slope intercept form is given by;

$y=mx+b$

where m is the slope

b is the y-intercept

x is the x-value on the coordinate plane

y is the y-value on the coordinate plane

$y-2x=2$ [Equation 1]

$y=2x+2$

This means that;

$m=2$

$y-intercept=2$

$-x=-y-1$ [Equation 2]

$y=x-1$

This means that;

$m=1$

$y-intercept=-1$

Both equations in the slope-intercept form are given by;

$\left\{\begin{array}{l}y=2x+2\\ y=x-1\end{array}\right.$

Let us find the y-value by assuming two values on the x-axis. Recall that two points are enough to give us a line. Given two values on the x-axis, we will use 1 and 2, what is y when x = 1? And what is y when x = 2?

The solution to these two questions should give us the lines of both equations.

Let us start with Equation 1,

$y=2x+2$.

Substitute 1 into the equation assuming x = 1,

$y=2\left(1\right)+2$

$y=4$

When $x=1$, $y=4$.

Substitute 2 into the equation assuming x = 2,

$y=2\left(2\right)+2$

$y=6$

When $x=2$, $y=6$.

We now have two points for Equation 1 to be plotted.

Graph of line y = 2x + 2 - StudySmarter Originals

The same will be done for the Equation 2,

$y=x-1$.

Substitute 1 into the equation assuming x = 1,

$y=1-1$

$y=0$

When $x=1$, $y=0$.

Substitute 2 into the equation assuming x = 2,

$y=2-1$

$y=1$

When $x=2$, $y=1$.

Let us plot these points and draw the line on the same coordinate plane.

Graph for equations y = 2x + 2 and y = x – 1, StudySmarter Originals

The point they both intercept is the solution for the problem, (–3, –4).

This means

$x=-3$

$y=-4$

Now we can evaluate this to see if this is true. Working with equations means that everything on the left side should be equal to what is on the right. Since we have two equations here, we will verify both. Let us start with the first one.

$y-2x=2$

We will substitute the values we just found into the equation

$-4-2(-3)=2$

Since both negative values are multiplying each other, the result becomes positive.

$-4+6=2$

$2=2$.

We do see here that the first equation is satisfied. We can go ahead to do the same with the second equation.

$-x=-y-1$

Substitute the values we just found into the equation

$-(-3)=-(-4)-1$

Negative values multiplying each other will result in positive.

$3=4-1$

$3=3$

We do realize here that the solution satisfies both equations, therefore, the solution is correct.