Nonlinear equations in one variable and systems of equations in two variables

The equation might have a square or higher power, factors, absolute value bars, a square root, or a variableA letter that stands for a number you don’t know yet, like the x in 5x + 7.Read the lesson in a denominatorThe bottom number of a fraction, like the 4 in 3/4.Read the lesson. Or it's a system of equationsTwo or more equations that all have to be true at the same time.Read the lesson pairing a line with a curve.

Read the last line first. It might say:

• “What is the positive solution to the given equation?”
• “How many distinct real solutions does the given equation have?”
• “What is the sum of the solutions to the given equation?”
• “At what point $(x, y)$ do the graphs of the equations in the given system intersect?”

Other questions ask for an unknown constantA number that doesn’t change, like the 7 in 5x + 7, or a letter such as p that stands for one fixed value.Read the lesson, or for one letter in terms of the others.

If the equation has only numbers and $x$, Desmos can check your work. Type y= and one side on line 1, then y= and the other side on line 2, and click where the graphs cross. It helps most with the quadraticAn expression whose biggest power is a square, like x squared plus 3x minus 4. formula, square roots and fractions. The solving by graphing lesson shows how.

If factors multiply to zero, one of them has to be zero.

If one side isn't zero yet, get it there first.

1. Start.

   $$(x+3)(x-8)+11=11$$

2. Subtract 11 from both sides.

   $$(x+3)(x-8)=0$$

3. Set each factor equal to 0.

   $$x=-3 \text{ or } x=8$$

When the same factor sits on both sides, move everything to one side and pull that factor out.

1. Start.

   $$(x+6)(x-2)=19(x+6)$$

2. Subtract $19(x+6)$ from both sides. The left side runs onto two lines.

   $$\begin{aligned}(x+6)(x-2)&\\-19(x+6)&=0\end{aligned}$$

3. Both parts contain $x+6$, so pull it out front.

   $$(x+6)\bigl((x-2)-19\bigr)=0$$

4. $-2-19=-21$.

   $$(x+6)(x-21)=0$$

5. Set each factor equal to 0.

   $$x=-6 \text{ or } x=21$$

A quadratic equation has an $x^2$ term. Try factoring first.

So $x=4$ or $x=7$. In Desmos, x^2-11x+28=0 on its own line draws a vertical line at each solution.

If you can't find the two numbers, use the quadratic formula. It works on every quadratic, and the reference sheetThe page of formulas you can open at any point in the Math section.Read the lesson doesn't print it.

Get every term on one side before you read off $a$, $b$, and $c$. $-3x^2-5x=-28$ becomes $-3x^2-5x+28=0$, so $c=28$, though the original shows $-28$.

1. Start.

   $$2x^2+6x-10=0$$

2. Divide every term by 2. No two whole numbers multiply to $-5$ and add to 3, so use the formula.

   $$x^2+3x-5=0$$

3. Read each number with its sign.

   $$a=1,\ b=3,\ c=-5$$

4. Work out $b^2-4ac$ first.

   $$3^2-4(1)(-5)$$

5. Minus times minus is plus.

   $$=9+20=29$$

6. Put $-b$ on top and $2a$ under the whole top. No whole number squares to 29, so $\sqrt{29}$ stays as it is.

   $$x=\frac{-3\pm\sqrt{29}}{2}$$

7. Take the plus sign.

   $$x=\frac{-3+\sqrt{29}}{2}$$

8. Take the minus sign.

   $$x=\frac{-3-\sqrt{29}}{2}$$

The part under the square root, $b^2-4ac$, is called the discriminantThe part under the square root in the quadratic formula, b squared minus 4ac. Positive means two real solutions, zero means one, negative means none.. You'll use it again to count solutions.

For $\frac{x^2}{16}=25$, multiply both sides by 16 to get $x^2=400$.

On a grid-inA question where you type your own answer instead of picking one of four.Read the lesson that asks for “one possible solution,” either one counts. The grid-in lesson covers what fits in the box.

Some questions put a quadratic in a rectangle story, such as a flat fee plus a charge for each square inch. To set one up:

1. Use a letter for the side you're asked for, such as $w$ for the width.
2. Write the other side with the same letter. “3 inches longer than the width” is $w+3$.
3. Take the flat fee off the total before you use the area.

Then solve, and drop the negative solution, since a length can't be negative. Give the side the question asks for.

Absolute value is distance from zero, so $|12|$ and $|-12|$ are both 12. That's why these equations split into two cases. Get the bars alone first.

1. Start.

   $$|x-7|+40=52$$

2. Subtract 40 to get the bars alone.

   $$|x-7|=12$$

3. The inside is 12 or $-12$.

   $$x-7=12 \text{ or } x-7=-12$$

4. Add 7 to both sides of each.

   $$x=19 \text{ or } x=-5$$

In Desmos, type y=abs(x-7)+40 and y=52, then zoom out until you see both crossings.

Which value solves $\sqrt{x+7}=x-5$?

• A. $2$
• B. $9$
• C. Both $2$ and $9$
• D. $\frac{1+\sqrt{129}}{2}$

1. The square root is alone, so square each whole side.

   $$x+7=(x-5)^2$$

2. Multiply out $(x-5)(x-5)$.

   $$x+7=x^2-10x+25$$

3. Subtract $x$ and 7 from both sides.

   $$0=x^2-11x+18$$

4. $-2$ and $-9$ multiply to 18 and add to $-11$.

   $$0=(x-2)(x-9)$$

5. Check 2 in the original: $3\ne-3$, so it fails.

   $$x=2:\ \sqrt{9}=3,\ 2-5=-3$$

6. Check 9: $4=4$, so it works.

   $$x=9:\ \sqrt{16}=4,\ 9-5=4$$

B is correct. Squaring hides signs, since $3^2$ and $(-3)^2$ are both 9. So 2 solves the squared equation even though the original says $3=-3$.

In Desmos, y=sqrt(x+7) and y=x-5 cross only at $x=9$.

Now try one with a letter in it.

Clear the fraction and solve, then throw out any answer that makes an original denominator zero.

1. Start. $x$ can't be 4, or the bottom is 0.

   $$\frac{x^2-16}{x-4}=7$$

2. Multiply both sides by $x-4$: $7(x-4)=7x-28$.

   $$x^2-16=7x-28$$

3. Subtract $7x$ and add 28 on both sides.

   $$x^2-7x+12=0$$

4. $-3$ and $-4$ multiply to 12 and add to $-7$.

   $$(x-3)(x-4)=0$$

5. Set each factor equal to 0.

   $$x=3 \text{ or } x=4$$

6. Throw out $x=4$, since it makes the original bottom zero.

   $$x=3$$

In Desmos, y=(x^2-16)/(x-4) and y=7 meet only at $x=3$. Type the fraction exactly as printed. After /, everything you type stays in the bottom until you press the right arrow.

The question reads something like “The given equation relates the positive numbers $q$, $r$, and $s$. Which equation correctly expresses $q$ in terms of $r$ and $s$?” To get one letter alone, undo each operation in reverse order.

Here's $F=\frac{kq}{d^2}$ solved for $d$, first with $F=12$, $k=3$, $q=16$.

1. Fill in the numbers: $kq=3\cdot16=48$.

   $$12=\frac{48}{d^2}$$

2. Multiply both sides by $d^2$.

   $$12d^2=48$$

3. Divide by 12.

   $$d^2=4$$

4. Take the square root. $d$ is positive, so keep 2.

   $$d=2$$

Now the same moves with letters.

1. Multiply both sides by $d^2$.

   $$Fd^2=kq$$

2. Divide by $F$.

   $$d^2=\frac{kq}{F}$$

3. Take the square root. The letters are positive, so keep the positive one.

   $$d=\sqrt{\frac{kq}{F}}$$

Desmos won't help here, since there are no numbers to graph.

With absolute value, the right side decides how many solutions there are. $|x-7|=12$ has two and $|x-7|=0$ has one, $x=7$. But $|x-7|=-5$ has none, because a distance is never negative.

“How many distinct real solutions” asks for a count. Distinct means a repeated value counts once. Real rules out values like the square root of $-4$, since no real number squared is negative.

List each factor's real solutions, then cross out repeats and count.

1. 4 is never 0, so none.

   $$4$$

2. No real solution here.

   $$x^2+9=0 \text{ means } x^2=-9$$

3. Both signs, so two.

   $$x^2-36=0 \text{ means } x=\pm6$$

4. The same value twice counts once.

   $$(x+2)^2=0 \text{ means } x=-2$$

5. Three distinct real solutions.

   $$-6,\ 6,\ -2$$

Higher powers work the same way. In $6(x^3+8)(x^4-81)=0$, $x^3=-8$ gives only $-2$, because a cube keeps its sign. $x^4=81$ gives $3$ and $-3$, and the 6 adds none, so that's three again.

For a quadratic, the discriminant counts for you. It's the part under $\pm\sqrt{\ }$.

So $x^2-12x+36=0$ has one solution even though it's a quadratic, because $(-12)^2-4(1)(36)=144-144=0$. It factors as $(x-6)(x-6)$, and a repeated 6 counts once.

In Desmos, type y= and the side with everything on it, then count where the curve crosses or touches the horizontal axisOne of the two lines a graph is measured against, one running across and one running up.. A touch counts once. Zoom out first, because a crossing off the screen still counts.

In the given equation, $c$ is a constant. The equation has no real solutions if $c\gt n$. What is the least possible value of $n$?

1. No real solutions means a negative discriminant. Here $a=1$ and $b=-30$.

   $$(-30)^2-4(1)(c)\lt0$$

2. $30^2=900$.

   $$900-4c\lt0$$

3. Add $4c$ to both sides.

   $$900\lt4c$$

4. Divide both sides by 4.

   $$225\lt c$$

So $n=225$. At $c=225$ itself the discriminant is 0, which gives one solution.

The letter can also sit in the middle term. In $r^2+qr=3r-40$, $q$ is an integer constant and the equation has no real solutions. What is the largest possible value of $q$?

1. Subtract $3r$ and add 40 on both sides.

   $$r^2+qr-3r+40=0$$

2. Group the $r$ terms, so $b=q-3$.

   $$r^2+(q-3)r+40=0$$

3. No real solutions, so the discriminant is negative.

   $$(q-3)^2-4(1)(40)\lt0$$

4. $4(40)=160$. Add 160 to both sides.

   $$(q-3)^2\lt160$$

5. Try whole numbers for $q-3$. 12 squared is under 160, and 13 squared isn't.

   $$12^2=144,\ \ 13^2=169$$

6. Add 3.

   $$q-3=12,\ \text{so}\ q=15$$

$x^2-11x+28=0$ has solutions 4 and 7. They add to 11 and multiply to 28. Here $b=-11$ and $c=28$, so the sum is $-b$ and the product is $c$. When $a$ isn't 1, divide both by $a$.

You get both without finding the solutions themselves.

An absolute value equation has its own shortcut to the sum.

1. Start.

   $$|5x-30|-4=21$$

2. Add 4 to get the bars alone.

   $$|5x-30|=25$$

3. The inside is 25 or $-25$.

   $$5x-30=\pm25$$

4. Add 30, then divide by 5.

   $$x=11 \text{ or } x=1$$

The sum is 12. Halfway between 1 and 11 is 6, the value that makes the inside zero, since $5(6)-30=0$. The two solutions always sit the same distance on either side of it, so next time you can skip the cases and double the middle value, $2\times6=12$.

A solution to a system is a point $(x, y)$ that makes both equations true, which is where the two graphs meet. Take $y=3x-6$ and $y=x^2-4x+6$.

1. Both equal $y$, so set them equal.

   $$3x-6=x^2-4x+6$$

2. Subtract $3x$ and add 6 on both sides.

   $$0=x^2-7x+12$$

3. $-3$ and $-4$ multiply to 12 and add to $-7$.

   $$0=(x-3)(x-4)$$

4. Set each factor equal to 0.

   $$x=3 \text{ or } x=4$$

5. Put $x=3$ into the line.

   $$y=3(3)-6=3$$

6. Put $x=4$ into the line.

   $$y=3(4)-6=6$$

7. Pair each $x$ with its own $y$.

   $$(3,3) \text{ and } (4,6)$$

On a graph, the solutions are where the two graphs cross or touch, so don't mistake an x-interceptThe value of x where a graph crosses the horizontal axis, so the output there is 0.Read the lesson for one. In Desmos, type y=3x-6 and y=x^2-4x+6 and click each crossing to read both coordinates.

A line and a parabolaThe U-shaped curve a quadratic draws, opening either up or down.Read the lesson can meet at two points, at one point, or not at all. Set them equal, and the discriminant of the result tells you which.

For $y=2$ and $y=x^2-6x+k$, which $k$ makes the graphs touch?

1. Both equal $y$, so set them equal.

   $$x^2-6x+k=2$$

2. Subtract 2. Then $a=1$, $b=-6$, $c=k-2$.

   $$x^2-6x+(k-2)=0$$

3. Touching once means the discriminant is 0.

   $$(-6)^2-4(1)(k-2)=0$$

4. $(-6)^2=36$, and $-4(k-2)=-4k+8$.

   $$36-4k+8=0$$

5. $36+8=44$. Add $4k$ to both sides.

   $$44=4k$$

6. Divide both sides by 4.

   $$k=11$$

On a graph, a curve that seems to touch may cross twice very close together, so go by the discriminant.

Now try one with an integer constant.

Before you start each one, decide what kind of equation it is. Check your answer against the original equation and the last line.