We've updated our
Privacy Policy effective December 15. Please read our updated Privacy Policy and tap

أدلة الدراسة > Prealgebra

Verifying Solutions to Equations in Two Variables

Learning Outcomes

  • Given a coordinate pair, determine whether it is a solution to a two variable equation

All the equations we solved so far have been equations with one variable. In almost every case, when we solved the equation we got exactly one solution. The process of solving an equation ended with a statement such as x=4x=4. Then we checked the solution by substituting back into the equation. Here’s an example of a linear equation in one variable, and its one solution.

3x+5=173x=12x=4\begin{array}{c}3x+5=17\hfill \\ \\ 3x=12\hfill \\ x=4\hfill \end{array} But equations can have more than one variable. Equations with two variables can be written in the general form Ax+By=CAx+By=C. An equation of this form is called a linear equation in two variables.  

Linear Equation

An equation of the form Ax+By=CAx+By=C, where AandBA\text{and}B are not both zero, is called a linear equation in two variables.
  Notice that the word "line" is in linear. Here is an example of a linear equation in two variables, xx and y:y\text{:} Ax+By=C\color{red}{A}x+\color{blue}{B}y=\color{green}{C} x+4y=8x+\color{blue}{4}y=\color{green}{8} A=1,B=4,C=8\color{red}{A=1},\color{blue}{B=4},\color{green}{C=8}   Is y=−5x+1y=-5x+1 a linear equation? It does not appear to be in the form Ax+By=CAx+By=C. But we could rewrite it in this form.
y=−5x+1y=-5x+1
Add 5x5x to both sides. y+5x=−5x+1+5xy+5x=-5x+1+5x
Simplify. y+5x=1y+5x=1
Use the Commutative Property to put it in Ax+By=CAx+By=C. Ax+By=C\color{red}{A}x+\color{blue}{B}y=C 5x+y=15x+y=1
By rewriting y=−5x+1y=-5x+1 as 5x+y=15x+y=1, we can see that it is a linear equation in two variables because it can be written in the form Ax+By=CAx+By=C. Linear equations in two variables have infinitely many solutions. For every number that is substituted for xx, there is a corresponding yy value. This pair of values is a solution to the linear equation and is represented by the ordered pair (x,y)\left(x,y\right). When we substitute these values of xx and yy into the equation, the result is a true statement because the value on the left side is equal to the value on the right side.  

Solution to a Linear Equation in Two Variables

An ordered pair (x,y)\left(x,y\right) is a solution to the linear equation Ax+By=CAx+By=C, if the equation is a true statement when the x-x\text{-} and y-valuesy\text{-values} of the ordered pair are substituted into the equation.
 

example

Determine which ordered pairs are solutions of the equation x+4y=8:x+4y=8\text{:} 1. (0,2)\left(0,2\right) 2. (2,−4)\left(2,-4\right) 3. (−4,3)\left(-4,3\right) Solution Substitute the x- andy-valuesx\text{- and}y\text{-values} from each ordered pair into the equation and determine if the result is a true statement.
1. (0,2)\left(0,2\right) 2. (2,−4)\left(2,-4\right) 3. (−4,3)\left(-4,3\right)
x=0,y=2x=\color{blue}{0}, y=\color{red}{2} x+4y=8x+4y=8 0+4⋅2=?8\color{blue}{0}+4\cdot\color{red}{2}\stackrel{?}{=}8 0+8=?80+8\stackrel{?}{=}8 8=8✓8=8\checkmark x=2,y=−4x=\color{blue}{2}, y=\color{red}{-4} x+4y=8x+4y=8 2+4(−4)=?8\color{blue}{2}+4(\color{red}{-4})\stackrel{?}{=}8 2+(−16)=?82+(-16)\stackrel{?}{=}8 −14=8-14=8 x=−4,y=3x=\color{blue}{-4}, y=\color{red}{3} x+4y=8x+4y=8 −4+4⋅3=?8\color{blue}{-4}+4\cdot\color{red}{3}\stackrel{?}{=}8 −4+12=?8-4+12\stackrel{?}{=}8 8=8✓8=8\checkmark
(0,2)\left(0,2\right) is a solution. (2,−4)\left(2,-4\right) is not a solution. (−4,3)\left(-4,3\right) is a solution.
 

try it

[ohm_question]146928[/ohm_question]
 

example

Determine which ordered pairs are solutions of the equation. y=5x−1:y=5x - 1\text{:} 1. (0,−1)\left(0,-1\right) 2. (1,4)\left(1,4\right) 3. (−2,−7)\left(-2,-7\right)

Answer: Solution Substitute the x-x\text{-} and y-valuesy\text{-values} from each ordered pair into the equation and determine if it results in a true statement.

1. (0,−1)\left(0,-1\right) 2. (1,4)\left(1,4\right) 3. (−2,−7)\left(-2,-7\right)
x=0,y=−1x=\color{blue}{0}, y=\color{red}{-1} y=5x−1y=5x-1 −1=?5(o)−1\color{red}{-1}\stackrel{?}{=}5(\color{blue}{o})-1 −1=?0−1-1\stackrel{?}{=}0-1 −1=−1✓-1=-1\checkmark x=1,y=4x=\color{blue}{1}, y=\color{red}{4} y=5x−1y=5x-1 4=?5(1)−1\color{red}{4}\stackrel{?}{=}5(\color{blue}{1})-1 4=?5−14\stackrel{?}{=}5-1 4=4✓4=4\checkmark x=−2,y=−7x=\color{blue}{-2}, y=\color{red}{-7} y=5x−1y=5x-1 −7=?5(−2)−1\color{red}{-7}\stackrel{?}{=}5(\color{blue}{-2})-1 −7=?−10−1-7\stackrel{?}{=}-10-1 −7=−11-7=-11
(0,−1)\left(0,-1\right) is a solution. (1,4)\left(1,4\right) is a solution. (−2,−7)\left(-2,-7\right) is not a solution.

 

try it

[ohm_question]146929[/ohm_question] [ohm_question]146941[/ohm_question]
 

Licenses & Attributions