How to solve problems in math with linear equations

Solving Linear Equations

A linear equation is important to technical jobs knowing that this is one of the indispensable tools when knowing the relationship between two variables. Engineers, accountants, economists, traders, stock marketers, and almost all professions require knowledge and understanding about linear equations. That is why this cannot be omitted in every curriculum. The future prediction relies based on how to read and interpret its meaning. Scientists use this tool to tell and interpret what was happening in their daily respective jobs. The graphical presentation using this tool is visually easy to understand. Without a linear equation, numerical alone might too abstract for some viewers. Ratio and proportion is the daily activity that involves linear equation such as wage per hour for company employees, mile per gallon ratio of car's engine, distance per minute we traveled, amount per item when we buy groceries, etc. There, we can adjust how to manage our time, monetary budget, and the like. So, let us take what it takes to learn some of its basics.

An equation of the form Ax + By + C = 0 is said to be linear if A, B, and C are constants. Constants refer to the numerical presentation, it is a number. The word linear means "line" so we are talking straight lines here. The Ax + By + C = 0 equation is in standard form and subject to some mathematical manipulation to attain a solution. Solving mathematical problems with linear equations is just as easy as playing puzzles. But this page talks about the manual pen-paper traditional school-based. Usually, the given unknown is "x and y", but is not limited to. Adding or subtracting both sides by what is to be eliminated is one of the easiest keys to followed by. If we want to see what it looks like, we prepared something for us to digest. The basic requirement to handle these problems is simply mastering collecting similar terms. Similar terms are Algebraic terms that have common literal coefficients including power. We combine all of them to become one term by means of either addition or subtraction. See the sample step guide below.

Sample Step Guide on How to Do It

Simplify the given equations by collecting similar terms.

$1.) {\color{Red} x}+2{\color{Red} x}=2$ , the "x" is common in this illustration. We use their numerical coefficients as the basis of adding or subtracting. So in this case, we add 1 and 2 as their numerical coefficients. So we have 3x = 2 in this equation, or x = 2/3. May the graph below clarify what's going on.

Graph x = 2/3

$2.)3{\color{Red} x}-6{\color{Red} x}+{\color{Red} x}=4$ , "x", is our common literal coefficient. So if we combine their numerical coefficients, we have "-2x = 4, or x = -2".

Graph of x = -2

$3.) 5{\color{Red} x}-2{\color{Blue} y}=5$ , so we have no literal coefficients in this illustration to collect. But we have its graph below.

Graph 5x-2y =5

$4.) 2-\frac{1}{2}{\color{Red} a}-\frac{3}{4}{\color{Red} a}-{\color{Red} a}=3b$ , So obviously, by the aid of colors, we can easily identify what is common. This is made just for similar terms collection and not for graphing although it's possible. Can you tell what is common? Click here!

$5.)4ab+3ab -ab =5$ , No colors but to test yourself if you can identify what are common. If you do then, Click here!

Now, we have basic ideas on how to collect similar terms. We will apply it now to actual problems in solving linear equations in one unknown. See an additional illustration of how to do it.

Solve for the value of x in every given equation. Assume that y = 2.

$1.) x+2y=5$ , so for this example, we don't have similar terms yet.

$x+2y-{\color{Blue} 2y}=5-{\color{Blue} 2y}$ , this is by subtracting 2y both sides of the equation.

$x=5-2y$ , this is now the value of "x". Now substituting y value which is given, then;

$x=5-2(y),@{\color{Blue} y=2}$

$x=5-2({\color{Blue} 2})$

$x=5-4$

$x=1$ , so this means that @ y= 2, x = 1 based on the given equation. See its graph below.

graph of x + 2y = 5

$2.) x+2y+4x =2x-5$ , notice that we have a literal coefficient "x" to collect as similar terms.

$x+2y-{\color{Blue} 2y}+4x-{\color{Red} 2x} =2x-5-{\color{Red} 2x}-{\color{Blue} 2y}$ , so this is by subtracting 2x and 2y both sides of the equation.

$3x=-5-2y$ , the result after subtracting 2x and 2y both sides of the equation. So, we need to eliminate 3 from the "3x" to find the value of x.

$(\frac{1}{{\color{DarkGreen} 3}}){\color{DarkGreen} 3}x=(-5-2y)(\frac{1}{3})$ , multiplying both sides by 1/3.

$x=\frac{-5-2y}{3}$ , now solving x @ y = 2, we have to substitute the value of given y.

$x=\frac{-5-2y}{3}, @{\color{Blue} y=2}$

$x=\frac{-5-2({\color{Blue} 2})}{3}$

$x=\frac{-5-4}{3}$

$x=\frac{-9}{3}$

$x=-3$ , so this means x = -3 @ y = 2 with respect to the given equation. See the graph below.

Graph of
$3x=-5-2y$

$3.) 5x=2x- 3y + 7$ , so 5x and 2x are similar terms to be collected.

$5x-{\color{Red} 2x}=2x-{\color{Red} 2x}- 3y + 7$ , this is by subtracting 2x from both sides of the equation to simply.

$3x=-3y + 7$ , now the result after subtracting 2x from both sides of the equation. We are ready to substitute y = 2.

$3x=-3y + 7, @{\color{Blue} y=2}$

$3x=-3({\color{Blue} 2}) + 7$

$3x=-6 + 7$

$3x=1$

$x=\frac{1}{3}$ , so x = 1/3 @ y = 2 from the given equation.

This time let's try solving for the value of y when x is given.

@ x = 4, solve for the value of y from the given equations below. Let's use the equation given from #1 above to compare how'd they go.

$4.) x+2y=5,@{\color{Red} x=4}$

$x-{\color{Red} x}+2y=5-{\color{Red} x}$ , subtracting x both sides of the equation to isolate y.

$2y=5-{\color{Red} x}$ , this is now the result after subtracting x both sides of the equation. Now let's isolate y.

$(\frac{1}{{\color{DarkGreen} 2}}){\color{DarkGreen} 2}y=({5-x})\frac{1}{2}$ , so multiply both sides by 1/2.

$y=\frac{5-x}{2}$ , after multiplying both sides by 1/2.

Now @ x = 4, let's solve the value of y.

$y=\frac{5-x}{2},@{\color{Red} x=4}$

$y=\frac{5-{\color{Red} 4}}{2}$

$y=\frac{1}{2}$ , so we have y = 1/2 @ x = 4 from the given equation.

Graph of x + 2y = 5

$5.) \frac{1}{2}{\color{Red} x}-2{\color{Red} x}+\frac{2}{3}{\color{Blue} y}=\frac{1}{3}{\color{Blue} y}-{\color{Red} x}+3, @{\color{Red} x=4}$ , x and y are highlighted in colors for clarity.

Now, let's collect similar terms to isolate y.

$\frac{1}{2}{\color{Red} x}-2{\color{Red} x}+\frac{2}{3}{\color{Blue} y}+{\color{Red} x}=\frac{1}{3}{\color{Blue} y}-{\color{Red} x}+3+{\color{Red} x}$

$-\frac{1}{2}{\color{Red} x}+\frac{2}{3}{\color{Blue} y}=\frac{1}{3}{\color{Blue} y}+3$ , this is now the result after adding both sides by x. Now let's work on y's.

$-\frac{1}{2}{\color{Red} x}+\frac{2}{3}{\color{Blue} y}-\frac{1}{3}{\color{Blue} y}=\frac{1}{3}{\color{Blue} y}+3-\frac{1}{3}{\color{Blue} y}$ , this is how it looks like when subtracting both sides by 1/3y.

$-\frac{1}{2}{\color{Red} x}+\frac{2}{3}{\color{Blue} y}-\frac{1}{3}{\color{Blue} y}=3$ , this is the detailed step-by-step so see the result after the operation.

$-\frac{1}{2}{\color{Red} x}+\frac{1}{3}{\color{Blue} y}=3$ , the result after the operation. Let's isolate y the process.

$-\frac{1}{2}{\color{Red} x}+\frac{1}{2}{\color{Red} x}+\frac{1}{3}{\color{Blue} y}=3+\frac{1}{2}{\color{Red} x}$ , now adding 1/2x to both sides.

$\frac{1}{3}{\color{Blue} y}=3+\frac{1}{2}{\color{Red} x}$ , the result after adding 1/2x to both sides.

${\color{DarkGreen} 3}(\frac{1}{3}{\color{Blue} y})=(3+\frac{1}{2}{\color{Red} x}){\color{DarkGreen} 3}$ , multiplying "3" to both sides of the equation.

$y{\color{Blue} }=9+\frac{3}{2}{\color{Red} x}$ , the result after multiplying "3" to both sides of the equation. Now let's substitute the value of x which is 4.

$y{\color{Blue} }=9+\frac{3}{2}{\color{Red} (4)}$

$y{\color{Blue} }=9+\frac{3}{{\color{DarkGreen} 2}}{\color{Red} ({\color{DarkGreen} 2}*2)}$ , canceled out the 2's.

$y{\color{Blue} }=9+3({\color{Red} 2})$

$y{\color{Blue} }=9+6$

$y{\color{Blue} }=15$ , so y = 15 @ x = 4 based on the given equation.

Graph of
$y{\color{Blue} }=9+\frac{3}{2}{\color{Red} x}$

Example Problems with Solutions

Solve for the value of x. Given y = -2.

$1.) 2{\color{Blue} y}+3{\color{Red} x}=4{\color{Red} x}-{\color{Blue} y}$

$2{\color{Blue} y}+3{\color{Red} x}-2{\color{Blue} y}=4{\color{Red} x}-{\color{Blue} y}-2{\color{Blue} y}$

$3{\color{Red} x}=4{\color{Red} x}-3{\color{Blue} y}$

$3{\color{Red} x}-4{\color{Red} x}=4{\color{Red} x}-3{\color{Blue} y}-4{\color{Red} x}$

$-{\color{Red} x}=-3{\color{Blue} y}$

${\color{Red} x}=3{\color{Blue} y}$ , so @ y = -2

${\color{Red} x}=3({\color{Blue} -2})$

${\color{Red} x}=-6$

So @ x = -6 , y = -2

$2.)\frac{3}{2}{\color{Red} x}+\frac{1}{4}{\color{Blue} y}+6=0$

$\frac{3}{2}{\color{Red} x}+\frac{1}{4}{\color{Blue} y}+6-\frac{1}{4}{\color{Blue} y}=-\frac{1}{4}{\color{Blue} y}$

$\frac{3}{2}{\color{Red} x}+6=-\frac{1}{4}{\color{Blue} y}$

$\frac{3}{2}{\color{Red} x}+6-6=-6 -\frac{1}{4}{\color{Blue} y}$

$\frac{3}{2}{\color{Red} x}=-6 -\frac{1}{4}{\color{Blue} y}$

${\color{DarkGreen} 2}(\frac{3}{{\color{DarkGreen} 2}}{\color{Red} x})=(-6 -\frac{1}{4}{\color{Blue} y}){\color{DarkGreen} 2}$

$3{\color{Red} x}=-12 -\frac{1}{2}{\color{Blue} y}$

$\frac{1}{{\color{DarkGreen} 3}}({\color{DarkGreen} 3}{\color{Red} x})=(-12 -\frac{1}{2}{\color{Blue} y})\frac{1}{3}$

${\color{Red} x}=-4 -\frac{1}{6}{\color{Blue} y}$ , @ y = -2

${\color{Red} x}=-4 -\frac{1}{6}{ (-{\color{Blue} 2})}$

${\color{Red} x}=-4 +\frac{1}{3}$

${\color{Red} x}=-\frac{11}{3}$

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

So @ x = -3(2/3) , y = -2

$3.)3{\color{Red} x}-2a{\color{Red} x}-1 =3{\color{Blue} y}+5$

${\color{Red} x}(3-2a)-1 =3{\color{Blue} y}+5$

${\color{Red} x}(3-2a)-1 +1=3{\color{Blue} y}+5 +1$

${\color{Red} x}(3-2a)=3{\color{Blue} y}+6$

$\frac{1}{({\color{DarkGreen} 3-2a})}[{\color{Red} x}({\color{DarkGreen} 3-2a})]=[3{\color{Blue} y}+6]\frac{1}{({\color{DarkGreen} 3-2a})}$

${\color{Red} x}=[3{\color{Blue} y}+6]\frac{1}{({\color{DarkGreen} 3-2a})}$ , @ y = -2

${\color{Red} x}=[3{\color{Blue} (-2)}+6]\frac{1}{({\color{DarkGreen} 3-2a})}$

${\color{Red} x}=[-6+6]\frac{1}{({\color{DarkGreen} 3-2a})}$

${\color{Red} x}=[0]\frac{1}{({\color{DarkGreen} 3-2a})}$

${\color{Red} x}=0$

So x = 0 @ y = -2

$4.) \frac{2}{3}{\color{Red} x}+5=\frac{1}{3}{\color{Red} x}-\frac{3}{4}{\color{Blue} y}$

$\frac{2}{3}{\color{Red} x}+5-\frac{1}{3}{\color{Red} x}=\frac{1}{3}{\color{Red} x}-\frac{3}{4}{\color{Blue} y}-\frac{1}{3}{\color{Red} x}$

$\frac{1}{3}{\color{Red} x}+5-{\color{DarkGreen} 5}=-\frac{3}{4}{\color{Blue} y}-{\color{DarkGreen} 5}$

$\frac{1}{3}{\color{Red} x}=-\frac{3}{4}{\color{Blue} y}-{\color{DarkGreen} 5}$

${\color{DarkGreen} 3}(\frac{1}{{\color{DarkGreen} 3}}{\color{Red} x})=(-\frac{3}{4}{\color{Blue} y}-5){\color{DarkGreen} 3}$

${\color{Red} x}=-\frac{9}{4}{\color{Blue} y}-15$ , @ y = -2

${\color{Red} x}=-\frac{9}{4}{\color{Blue} (-2)}-15$

${\color{Red} x}=\frac{18}{4}-15$

${\color{Red} x}=\frac{-21}{2}$

${\color{Red} x}=-10\frac{1}{2}$

So @ y= -2 , x = -10 1/2

$5.)2{\color{Red} x} +3y{\color{Red} x}=10$

${\color{Red} x}(2 +3y)=10$

$\frac{1}{({\color{DarkGreen} 2 +3y})}[{\color{Red} x}({\color{DarkGreen} 2 +3y})]=(10)\frac{1}{(2 +3y)}$

${\color{Red} x}=(10)\frac{1}{(2 +3y)}$

${\color{Red} x}=\frac{10}{2 +3y}$ ,@ y = - 2

${\color{Red} x}=\frac{10}{2 +3(-{\color{Blue} 2})}$

${\color{Red} x}=\frac{10}{2 -6}$

${\color{Red} x}=\frac{10}{-4}$

${\color{Red} x}=-\frac{5}{2}=-2\frac{1}{2}$

So @ y= -2 , x = -2 1/2

Problems with Partial Solutions

**Given the equations below see (numbers 1-5), Solve for the value of "what is asked" based on the given.**

$1.)2=3x+4y$ , @ x = -1

$2{\color{DarkGreen} -3x}=3x{\color{DarkGreen} -3x}+4y$

$2{\color{DarkGreen} -3x}=4y$

$4y=2{\color{DarkGreen} -3x}$

$\frac{1}{{\color{DarkGreen} 4}}({\color{DarkGreen} 4}y)=(2{\color{DarkGreen} -3x})\frac{1}{4}$

$y=(2{\color{DarkGreen} -3x})\frac{1}{4}$

$y=\frac{2{-3x}}{4}$ , @ x = -1

y = click to put your answer

$2.) 3{\color{Red} x}-4{\color{Red} x}y = {\color{Red} x} - {\color{Red} x}y$

$3{\color{Red} x}-4{\color{Red} x}y{\color{DarkGreen} -x+{ x}y}={\color{Red} x}-{\color{Red} x}y{\color{DarkGreen} -x+{ x}y}$

$3{\color{Red} x}-4{\color{Red} x}y{\color{DarkGreen} -x+{ x}y}=0$

$2{\color{Red} x}-4{\color{Red} x}y{+{ {\color{Red} x}}y}=0$

${\color{Red} x}(2{}-4{}y{+{ {}}y})=0$

${\color{Red} x}(2{}-3{}y)=0$

$\frac{1}{({\color{DarkGreen} 2-3y})}[{\color{Red} x}({\color{DarkGreen} 2-3y})]=(0)\frac{1}{(2-3y)}$

${\color{Red} x}=(0)\frac{1}{(2-3y)}$ @ y = -1

x = click to put your answer

$3.) 2{\color{Red} x}y +5b{\color{Red} x} = 2-4y , @{\color{Blue} y}=3$

${\color{Red} x}(2y +5b) = 2-4y$

$\frac{1}{({\color{DarkGreen} 2y +5b})}[{\color{Red} x}({\color{DarkGreen} 2y +5b})] = (2-4y)\frac{1}{(2y +5b)}$

${\color{Red} x} = \frac{2-4{\color{Blue} y}}{2{\color{Blue} y} +5b}$

${\color{Red} x}=$ Click to put your answer

$4.)\frac{2}{3}x+\frac{3}{4}y=1, @ {\color{Red} x}=-1$

${\color{DarkGreen} \frac{2}{3}x}+\frac{3}{4}y-{\color{DarkGreen} \frac{2}{3}x}=1-\frac{2}{3}x$

$\frac{3}{4}y=1-\frac{2}{3}x$

$\frac{{\color{DarkGreen} 4}}{{\color{DarkGreen} 3}}(\frac{{\color{DarkGreen} 3}}{{\color{DarkGreen} 4}}y)=(1-\frac{2}{3}x)\frac{4}{3}$

$y=(1-\frac{2}{3}x)\frac{4}{3}$

$y=(\frac{4}{3}-\frac{8}{9}{\color{Red} x})$

$y=$ Click to put your answer

$5.) 3xy-4=6, @{\color{Red} x}=5$

$3xy-4+4=6+4$

$3xy=10$

$\frac{1}{{\color{DarkGreen} 3x}}({\color{DarkGreen} 3x}y)=(10)\frac{1}{3x}$

$y=\frac{10}{3{\color{Red} x}}$

$y=$ Click to put your answer