Systems of two linear equations in two variables

A system of equationsTwo or more equations that all have to be true at the same time. is two equations about the same two unknowns, usually $x$ and $y$. On a graph each one is a straight line. A solution is one pair of numbers that makes both equations true.

Read the last line first. It might say:

• “The solution to the given system of equations is $(x, y)$. What is the value of $x$?” Some ask for $x+y$ instead.
• “What system of linear equations is represented by the lines shown?”
• “Which system of equations describes this situation?”
• “Which of the following systems of linear equations has no solution?”
• “In the given system of equations, $p$ is a 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. If the system has no solution, what is the value of $p$?” Here $p$ stands for one number you haven't been told, and it's the answer.

If one letter is already alone, put it into the other equation. If not, make one letter cancel, or let Desmos solve it.

The solution to the given system of equations is $(x, y)$. What is the value of $x$?

1. The first equation says $y$ is $2x-1$. Put that into the second equation, in parentheses.

   $$3x+4(2x-1)=29$$

2. Multiply both terms in the parentheses by 4.

   $$3x+8x-4=29$$

3. Combine $3x$ and $8x$.

   $$11x-4=29$$

4. Add 4 to both sides.

   $$11x=33$$

5. Divide both sides by 11.

   $$x=3$$

6. Put 3 into the first equation to get $y$.

   $$y=2(3)-1=5$$

7. Check in the second equation: $9+20=29$.

   $$3(3)+4(5)=29$$

The question asks for $x$, so the answer is 3. Once $y$ is gone, you're solving an equation with one letter, which the one-variable lesson covers.

The expression has to go into the other equation. If you put $2x-1$ back into $y=2x-1$, you get $2x-1=2x-1$, which is true for every $x$ and doesn't tell you anything.

When both equations look like $ax+by=c$, you can add them so one letter drops out. The number in front of a letter is its coefficientThe number multiplying a letter, like the 5 in 5x.Read the lesson. A letter cancels when its two coefficients are opposites, like $12y$ and $-12y$.

1. Multiply every term of the first equation by 3.

   $$9x+12y=345$$

2. Multiply every term of the second by 4. Now the $y$ terms are $12y$ and $-12y$.

   $$28x-12y=284$$

3. Add the equations: left side plus left side, right side plus right side.

   $$(9x+28x)+(12y-12y)=345+284$$

4. $12y-12y$ is 0, so $y$ is gone.

   $$37x=629$$

5. Divide both sides by 37.

   $$x=17$$

6. Put 17 into the second original equation.

   $$7(17)-3y=71$$

7. Subtract 119 from both sides.

   $$-3y=-48$$

8. Divide both sides by $-3$.

   $$y=16$$

9. Check in the first equation: $51+64=115$.

   $$3(17)+4(16)=115$$

You don't always need to multiply first. In $3x+4y=26$ and $8x+4y=51$, both $y$ terms are already $+4y$.

Sometimes the last line asks for $x+y$ or $x-y$, and you never need $x$ or $y$ alone. Look for the same two coefficients in swapped places, as in the system from the first section:

What is the value of $x+y$?

1. Add the equations.

   $$(3x+5x)+(5y+3y)=14+18$$

2. Both letters now have 8 in front.

   $$8x+8y=32$$

3. Divide every term by 8.

   $$x+y=4$$

The solution is $(3,1)$, and $3+1=4$, so it checks. Subtracting the first equation from the second gives $2x-2y=4$ instead, so $x-y=2$.

Adding and subtracting equations only gives sums and differences. For a product like $xy$, solve for both letters and multiply, which here is $3\times1=3$.

If the last line asks for a multiple like $11x$, watch for it on the way. Solving $y=2x-1$ and $3x+4y=29$ reaches $11x=33$ a step before $x=3$, so the answer is 33. If the multiple never shows up, find $x$ and multiply.

Here's one to try.

Name each unknown in words before you write anything: “$w$ = water bottles, $h$ = hoodies.” Then turn each fact into its own equation. Often one fact is a count and the other is an amount of money.

A school club sold 58 water bottles and hoodies at a fundraiser. Water bottles cost $8, hoodies cost $14, and the club took in $590 in all.

1. Get $w$ alone in the count equation by subtracting $h$.

   $$w=58-h$$

2. Put it into the money equation, in parentheses.

   $$8(58-h)+14h=590$$

3. Multiply both terms by 8.

   $$464-8h+14h=590$$

4. Combine $-8h$ and $14h$.

   $$464+6h=590$$

5. Subtract 464 from both sides.

   $$6h=126$$

6. Divide both sides by 6.

   $$h=21$$

7. Find $w$.

   $$w=58-21=37$$

8. Check the money: $296+294=590$.

   $$8(37)+14(21)=590$$

That's 37 water bottles and 21 hoodies.

“Times as many” gets turned around a lot. Say a farm stand sold 5 times as many ears of corn as melons.

If the stand sold 240 items in all, the second equation is $C+M=240$, and $5M+M=240$ gives 40 melons.

If the question only asks which system describes the story, stop once both equations match a choice. Here's a word problem to try.

+−100%Reset⤢

At Rafiq’s museum, adult admission cost $8, youth admission cost $5, and 13 more adults than youths visited.

How many youth visitors came to the museum that day?

1. A

   78

2. B

   75

3. C

   70

4. D

   62

Show the answer

• A

  78

  This results from writing the comparison backward as $x=y+13$, which treats the number of youths as 13 greater than the number of adults and produces $x=78$.

• B

  75

  This is the adult count, since $y=62+13=75$. The question asks for the number of youth visitors.

• C

  70

  This results from calculating $910\div(5+8)=70$, which ignores that 13 additional adults each paid $\$8$.

• D

  62

  Correct answer

  Let $x$ be the number of youth visitors. Since there were 13 more adults, the revenue equationTwo expressions set equal to each other, like 5x + 7 = 22.Read the lesson is $5x+8(x+13)=910$. Solving gives $13x=806$, so $x=62$.

Solution

Let $x$ be the number of youth visitors. Then the number of adult visitors is $x+13$.

Use the admission prices and total revenue to write $5x+8(x+13)=910$. Distribute and combine like termsParts of an expression with the same letter raised to the same power, like 5x and 2x, which you can add together.Read the lesson:

$5x+8x+104=910$

$13x=806$

$x=62$

Therefore, 62 youth visitors came to the museum that day.

Report a problem

Fractions make it hard to match coefficients by hand. Multiply each whole equation by a number all its bottoms go into.

When a letter sits on the right side, move it left first.

1. Start from the equation.

   $$\frac{5}{6}s=\frac{3}{8}r+2$$

2. Subtract $\frac{3}{8}r$ from both sides, so both letters are on the left.

   $$\frac{5}{6}s-\frac{3}{8}r=2$$

3. Multiply every term by 24, since 6 and 8 both go into 24.

   $$20s-9r=48$$

A word problem can hand you the fractions. “One-fourth of $r$ exceeds one-sixth of $d$ by $\frac{2}{3}$” works like this:

1. “Exceeds by” means the first amount minus the second.

   $$\frac{r}{4}-\frac{d}{6}=\frac{2}{3}$$

2. Multiply every term by 12.

   $$3r-2d=8$$

The choices may show only the cleared form, so you'd look for $3r-2d=8$.

In Desmos you can skip the cleanup and type the fractions, pressing the right arrow after each bottom number. For $\frac{3x}{4}+\frac{y}{6}=\frac{5}{12}$, type 3/4 → x+1/6 → y=5/12, where → is the right arrow. Check that both letters sit outside the fractions.

On a graph, the solution is the point where the lines cross.

“What system of linear equations is represented by the lines shown?” asks for both equations. Find each line's slopeHow steep a line is, measured by how much it goes up for each unit it moves to the right.Read the lesson and y-interceptThe output when the input is 0, which is where a graph crosses the vertical axis.Read the lesson from its own two labeled points. For a line through $(0,9)$ and $(7,23)$:

1. Change in $y$ over change in $x$.

   $$m=\frac{23-9}{7-0}=2$$

2. The line crosses the vertical axis at $(0,9)$, so $b=9$.

   $$y=2x+9$$

Do the other line the same way. If the choices look like $ax+by=c$, move the $x$ term over to compare: $y=2x+9$ is $-2x+y=9$. The linear functions lesson shows how to get an equation from two points.

Typing both equations into Desmos takes about ten seconds, which is worth it when there are fractions, messy numbers, or letters on both sides. If one equation already gives you a value, like $y=23$, or one subtraction finishes the work, it's quicker by hand.

If you don't see a crossing, zoom out before you decide there isn't one. The coordinates show as decimals. A value like 3.5 goes into a grid-inA question where you type your own answer instead of picking one of four.Read the lesson as it is. For a long decimal, finish by hand to get the exact fraction. The solving by graphing lesson has more.

Put both equations in $y=mx+b$ form, then compare the slopes, and after that the y-intercepts.

When both equations look like $ax+by=c$, multiply one so its left side matches the other's, then compare the right sides.

1. Triple every term of the first equation. Now the left sides match.

   $$9x+21y=60$$

2. The right sides don't. The lines are parallel and never meet, so there's no solution.

   $$60\ne50$$

Some choices have a letter in every equation. Compare $y=ax-3$ with $6y=6ax+12$.

1. Divide every term of the second equation by 6.

   $$y=ax+2$$

2. Same slope, $a$, and different y-intercepts. No solution, whatever $a$ is.

   $$-3\ne2$$

A question can also add a third line to a system you've solved. Say $y=2x+1$ and $y=-x+7$ cross at $(2,5)$, and the third line is $y=x+3$. Put in $x=2$: $2+3=5$, so the third line goes through $(2,5)$ too, and the three equations have one solution. If the third line were $y=x+4$, it would miss $(2,5)$, and there'd be no solution.

Here the question tells you the system has no solution and asks for the letter. Make the left sides match, then check that the right sides don't.

1. Triple every term of the first equation, so its $x$ term matches the $6x$.

   $$6x+15y=27$$

2. For no solution, the $y$ terms have to match too.

   $$p=15$$

3. Check the right sides. They're different, so the lines are parallel.

   $$27\ne4$$

Don't skip the last row. If the second equation had been $6x+py=27$, then $p=15$ would make it the same line, with infinitely many solutions.

The harder version puts a letter on the right side, so the printed numbers aren't the real ones.

In the given system of equations, $p$ is a constant. If the system has no solution, what is the value of $p$?

1. Subtract $5y$ from both sides of the first equation.

   $$6x+6y=-7$$

2. Multiply every term of the second equation by $-3$, so its $y$ term becomes $6y$.

   $$-3px+6y=-9$$

3. For no solution, the $x$ terms have to match too.

   $$-3p=6$$

4. Divide both sides by $-3$.

   $$p=-2$$

5. Check: this is the second equation times $-3$ with $p=-2$. $-9\ne-7$, so there's no solution.

   $$6x+6y=-9$$

Negative and fraction answers are normal here, so don't change your work because of one.

Now try one where the constant is squared.

Before each one, write what the last line asks for. When you finish, check your answer against both equations.