How to Solve Equations With Variables on Both Sides

How to Solve Equations With Variables on Both Sides – Sometimes, the unknown quantity will appear on both sides of an equation. This is where the properties learned in 5.1 and 5.2 come in handy. A quantity with a variable can be treated just like a quantity without variables — a quantity with a variable follows all the rules learned in the last two sections. For example, we can add a quantity with a variable to both sides without changing the equation or the values that make it true:

15 – x = 4x
15 – x + x = 4x + x
15 + 0x = 5x
15 = 5x
3 = x
x = 3

Table of Contents

Method 1: Solving Equations with One Variable on Both Sides

1Apply the distributive property, if necessary. The distributive property states that .This rule allows you to cancel out parentheses by multiplying each term in the parentheses by the number outside the parentheses.
- For example, if your equation is $2(10-2x)=4(2x+2)$ , use the distributive property to multiply the terms in parentheses by the number outside the parentheses:
  $2(10-2x)=4(2x+2)$
  $20-4x=8x+8$
Cancel the variable on one side of the equation. To cancel the variable, complete the opposite operation as stated in the equation. For example, if the term is subtracted in the equation, cancel it by adding. If the term is added in the equation, cancel it by subtracting. It is usually easiest to cancel the variable with the smaller coefficient.
- For example, in the equation $20-4x=8x+8$ , cancel the term $-4x$ by adding $4x$ :
  $20-4x+4x=8x+8$ .
Keep the equation balanced. Whatever you do to one side of the equation, you must do to the other side as well. So if you add or subtract to cancel the variable on one side of the equation, you must add or subtract to the other side as well.
- For example, if you added $4x$ on one side of the equation to cancel the variable, you must also add $4x$ to the other side of the equation:
  $20-4x+4x=8x+8+4x$
Simplify the equation by combining like terms. You should now have the variable on one side of the equation.
- For example:
  $20-4x+4x=8x+8+4x$
  $20=12x+8$
Move the constants to one side of the equation, if necessary. You want the variable term on one side, and the constant on the other side. To move the constant to one side, add or subtract from each side of the equation to cancel the term on one side.
- For example, to cancel the $+8$ constant on the variable side, subtract 8 from both sides of the equation:
  $20=12x+8$
  $20-8=12x+8-8$
  $12=12x$
Cancel the variable’s coefficient. To do this, perform the operation opposite from the one denoted in the equation. Usually this will mean dividing to cancel a coefficient being multiplied by a variable.^[4]Remember that whatever you do to one side of the equation, you must do to the other side of the equation as well.
- For example, to cancel out the coefficient 12 from the equation, you would divide each side of the equation by 12:
  $12=12x$
  ${\frac {12}{12}}={\frac {12x}{12}}$
  $1=x$
Check your work. To make sure your answer is correct, substitute your solution back into the original equation. If the equation is true, your answer is correct.
- For example, if $1=x$ , substitute 1 for the variable in the equation and calculate:
  $2(10-2x)=4(2x+2)$
  $2(10-2(1))=4(2(1)+2)$
  $2(10-2)=4(2+2)$
  $20-4=8+8$
  $16=16$

Method 2: Solving System Equations with Two Variables

1Isolate a variable in one equation. This might already be done. If not, use the rules of algebra to isolate the variable on one side of the equation. Remember that whatever you do to one side of the equation, you must do to the other side.
- For example, for the equation $y+1=x-1$ , to isolate the $y$ variable, you would subtract 1 from both sides:
  $y+1=x-1$
  $y+1-1=x-1-1$
  $y=x-2$
Substitute the value of the isolated variable into the other equation. Make sure you substitute the entire expression for the variable. This will give you an equation with only one variable, allowing you to solve for the variable.
- For example, if your first equation is $2x=20-2y$ , and you determined $y=x-2$ in the second equation, you would substitute $x-2$ for $y$ in the first equation:
  $2x=20-2y$
  $2x=20-2(x-2)$
Solve for the variable. To do this, move the variable to one side of the equation. Then, move the constants to one side of the equation. Then, isolate the variable using multiplication or division.
- For example:
  $2x=20-2(x-2)$
  $2x=20-2x+4$
  $2x=24-2x$
  $2x+2x=24-2x+2x$
  $4x=24$
  ${\frac {4x}{4}}={\frac {24}{4}}$
  $x=6$
Solve for the remaining variable. To do this, plug the value of the variable you already solved into one of the equations. This will give you an equation with only one variable. Solve for the variable using the rules of algebra. You can use either equation to solve for the remaining variable.
- For example, if you found that $x=6$ , you can substitute 6 for $x$ in the second equation:
  $y=x-2$
  $y=(6)-2$
  $y=4$
Check your work. Plug the values for both variables into one of the equations. If the equation is true, your solutions are correct.
- For example, if you found that $x=6$ and $y=4$ , plug these back into the original equation and solve:
  $2x=20-2y$
  $2(6)=20-2(4)$
  $12=20-8$
  $12=12$

Method 3: Solving Example Problems

Try this problem using the distributive property with one variable: .
- Use the distributive property to cancel the parentheses:
  $5(x+4)=6x-5$
  $5x+20=6x-5$
- Cancel the $5x$ on the left side of the equation by subtracting $5x$ from both sides:
  $5x+20=6x-5$
  $5x+20-5x=6x-5-5x$
  $20=x-5$
- Isolate the variable by adding 5 to each side of the equation:
  $20=x-5$
  $20+5=x-5+5$
  $25=x$
Try this problem involving a fraction: .
- Remove the fraction. To do this, multiply each side of the equation by the fraction’s denominator:
  $-7+3x={\frac {7-x}{2}}$
  $2(-7+3x)=2({\frac {7-x}{2}})$
  $-14+6x=7-x$
- Cancel the $-x$ on the right side of the equation by adding $x$ to each side of the equation:
  $-14+6x=7-x$
  $-14+6x+x=7-x+x$
  $-14+7x=7$
- Move the constants to one side of the equation by adding 14 to each side:
  $-14+7x=7$
  $-14+7x+14=7+14$
  $7x=21$
- Cancel the coefficient by dividing each side of the equation by 7:
  $7x=21$
  ${\frac {7x}{7}}={\frac {21}{7}}$
  $x=3$
Try solving this system of equations:
- Isolate the $y$ variable in the second equation:
  $9y=9x+27$
  $9y=9(x+3)$
  ${\frac {9y}{9}}={\frac {9(x+3)}{9}}$
  $y=x+3$
- Plug in $x+3$ for $y$ in the first equation:
  $9x+15=12y$
  $9x+15=12(x+3)$
- Use the distributive property to cancel the parentheses:
  $9x+15=12x+36$
- Cancel the variable on the left side of the equation by subtracting $9x$ from each side:
  $9x+15=12x+36$
  $9x+15-9x=12x+36-9x$
  $15=3x+36$
- Move the constants to one side by subtracting 36 from each side:
  $15=3x+36$
  $15-36=3x+36-36$
  $-21=3x$
- Cancel the coefficient by dividing each side by 3:
  $-21=3x$
  ${\frac {-21}{3}}={\frac {3x}{3}}$
  $-7=x$
- Solve for $y$ by plugging the value of $x$ into either equation:
  $9y=9x+27$
  } $9y=9(-7)+27$
  $9y=-63+27$
  $9y=-36$
  ${\frac {9y}{9}}={\frac {-36}{9}}$
  $y=-4$

Conclusion

Solving equations can be tough, especially if you’ve forgotten or have trouble understanding the tools at your disposal. One of those tools is the subtraction property of equality, and it lets you subtract the same number from both sides of an equation. Watch the video to see it in action!

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