Linear and Nonlinear Expressions

Linear and Nonlinear Expressions exercise

• Identify if an equation is linear and nonlinear.

  Hints

  Linear equations contain variables raised to a power of $1$.

  Each equation which is not linear is nonlinear.

  $y=x^2+1$ is an example of a nonlinear equation.

  Keep in mind: instead of writing $x^1$, we usually just write $x$.

  Solution

  Linear equations contain variables raised to a power of $1$.

  In other words: An equation is not linear and thus nonlinear if the variable is in the denominator, or squared, or cubed, or raised to another power greater than $1$. Another example for a nonlinear equation is given by $y$ equals the square root of $x$.

  So, now we are in the position to detect all linear equations. Those are:

  • $y=6x+3$
  • $y=3x-1$

  $ $

  All other equations are nonlinear:

  • $y=\frac2x$ as well as $y=\frac5x$ are rational equations.
  • $y=x^2$ is a square equation.
  • $y=-4x^3+8$ is a cube equation.
  • $y=\sqrt x$ is a root equation.

• Identify which graphs are linear and which are nonlinear.

  Hints

  The graph of a linear equation is a straight line.

  All variables in a linear equation are raised to the power of $1$.

  Just two graphs show linear equations.

  Solution

  How can you decide if an equation is linear or nonlinear just by looking at its graph?

  The graph of a linear equation is always a straight line.

  If the graph is not a straight line, its corresponding equation must be nonlinear.

  The following are linear equations:

  • $y=6x+3$
  • $y=3x-1$

• Find the nonlinear terms in the equations.

  Hints

  First, look at the exponents of the variables: if any exponent is greater than $1$, the equation can't be linear.

  An equation with any variable under a radical or in a denominator is nonlinear.

  Mofa found three possible routes that are linear.

  Solution

  Mofa wants to avoid further traffic tickets, so he plans his routes and writes down some equations. Unfortunately there are still some nonlinear equations. Those routes will lead to big trouble, so he has to exclude them.

  Keep in mind: All variables in linear equations are raised to the power of $1$, which is usually not written down.

  • $y+x=5$ Sure, this is a linear equation.
  • $y=\frac 5x$ is a rational equation. The variable $x$ appears in the denominator.
  • $y=\frac{1}{5} x$ This looks quite similar to the equation before. But this is also a linear equation.
  • $y=x^2-5$ Here the variable $x$ is squared. Thus this is a quadratic equation and therefore not a linear equation at all.
  • $y=\sqrt{x}+5$ is a radical equation.
  • $y=2x+5$ is a linear equation.

• Determine which factored equations are linear.

  Hints

  First establish each equation in the form $y=...$.

  Each linear equation is given by $y=mx+b$.

  All variables in linear equations are raised to the power of $1$.

  Use the distributive property to multiply, $2(x-2)=(2)(x)-(2)(2)=2x-4$.

  For multiplying two binomials you use the FOIL method. Let's have a look at the following example: $(x-1)(x+2)$.

  • Multiply the First: $x\times x=x^2.
  • Next, multiply the Outer: $x\times 2=2x$.
  • Now multiply the Inner: $(-1)\times x=-x$.
  • Last, multiply the Last: $(-1)\times 2=-2$.

  Finally, add all terms and combine the like terms:

  $(x-1)(x+2)=x^2+2x-x-2=x^2+x-2$.

  Solution

  We can write each linear equation in the form $y=mx+b$. Otherwise the equation isn't linear.

  $~$

  Route 1

  $y=2(x+1)-x=2x+2-x=2x-x+2=x+2$ $~~~~~$✓

  $~$

  Route 2

  $y=(2x+1)\times x=(2x)(x)+x=2x^2+x$

  This is a quadratic equation.

  $~$

  Route 3

  $y^2=2(x+1)-x$

  The right side is the same as that of route number $1$. Finally, we have to take the square root to get

  $y=\sqrt{x+2}$

  This is a radical equation.

  $~$

  Route 4

  $2\times x\times y=4$

  Dividing by $2$ and $x$ we get $y=\frac2x$.

  This is a rational equation.

  $~$

  Route 5

  $2x+y=5$

  Subtracting $2x$ and arranging the terms in the right order we get

  $y=-2x+5$ $~~~~~$✓

  $~$

  Route 6

  $y=(x+1)(x-2)$

  Using FOIL-multiplication we get

  $y=(x+1)(x-2)=x^2-2x+x-2=x^2-x-2$

  a quadratic equation.

• Determine which statements are true.

  Hints

  Here is a parabola. The corresponding equation is quadratic: $y=2x^2$.

  This is an example of a linear equation: $y=2x+4$.

  Solution

  Linear equations contain variables raised to the power of $1$.

  All other equations are nonlinear. When written in simplified form, here are some things that make an equation not linear:

  • having a variable in the denominator.
  • having a variable under a radical.
  • having a variable squared or cubed.

  If a linear equation is graphed, you see a straight line.

• Find the corresponding expression and state if it is linear or nonlinear.

  Hints

  Each time assign the variable $x$ to the unknown number first.

  To multiply two binomials use the FOIL-multiplication

  $(-2x+8)(x+10)$

  $=(-2x)(x)-(2x)(10)+(8)(x)+(8)(10)$

  $=-2x^2-20x+8x+80$

  $=-2x^2-12x+80$

  Use the distributive property to get

  $2(3x+6)=(2)(3x)+(2)(6)=6x+12$.

  Solution

  $y$ is the product of $7$ and an unknown number.

  $7$ times an unknown number can be written as $7x$. So we get

  $y=7x$,

  which is a linear equation.

  $y$ is equal to $5$ multiplied by the reciprocal of the quotient of an unknown number and $12$.

  To get the reciprocal you switch the numerator and denominator. With this we can establish the equation as follows:

  $y=5\times \frac{12}x=\frac{60}x$

  This is a rational equation and thus nonlinear.

  $y$ is given as the quantity eight minus twice an unknown number, multiplied by the sum of this unknown and $10$.

  Let's do this step by step

  • eight minus twice an unknown number is given by the binomial $8-2x=-2x+8$.
  • the sum of this unknown and $10$ can be written as $x+10$

  Now we are in the position to multiply those binomials using FOIL:

  $y=(-2x+8)(x+10)=(-2x)(x)-(2x)(10)+(8)(x)+(8)(10)$ $=-2x^2-20x+8x+80=-2x^2-12x+80$,

  which is quadratic and thus nonlinear.

  $y$ is the sum of four consecutive numbers, the smallest one represented by $x$.

  The four consecutive numbers are given as follows

  • $x$
  • $x+1$
  • $x+2$
  • $x+3$

  Adding those numbers leads to $y=x+x+1+x+2+x+3=x+x+x+x+1+2+3=4x+6$

  This is a linear equation.

  $y$ is given by the half of the product of a number multiplied by itself three times.

  Multiplying a number by itself can be written with exponents. So a number multiplied itself three times looks like

  $x\times x\times x=x^3$

  We get $y=\frac12x^3$,

  which is a nonlinear equation.

  Just one more example and we're done:

  Let $y$ be twice the sum of $6$ and $3$ times an unknown number.

  First, we establish the sum $3x+6$. Twice this sum is $2(3x+6)$. Putting all this together we can conclude

  $y=6x+12$,

  which is a linear equation.