Q&A for How to Solve Simultaneous Equations Using Substitution Method

Not really. Here's the simplest example possible: let's say x + y = 3 and x - y = 1. Solve the second equation for x by adding y to both sides: (x - y) + y = 1 + y. So x = 1 + y. Take that value of x, and substitute it into the first equation given above (x + y = 3). With that substitution the first equation becomes (1+y) + y = 3. That means 1 + 2y = 3. Subtract 1 from each side: 2y = 2. So y = 1. Substitute that value of y into either of the two original equations, and you'll get x = 2.

Yes, it's solvable. Take the second equation, and subtract 5x from both sides: y = (30 - 5x). Plug that value of y back into the other equation: 3x + 4(30 - 5x) = 52. So (3x + 120 - 20x) = 52, and (-17x + 120) = 52. Then (-17x) = -68, and x = 4. Plug that value of x into either of the original equations: 3(4) + 4y = 52, so 12 + 4y = 52, and 4y = 40, so that y = 10. Check your work by plugging the found values of x and y into either of the original equations.

It depends on what kind of equation you have. One skill you should develop is knowing when to use what so you can manage your time wisely.

There's no hard-and-fast rule. You just have to analyze the situation and decide which method would be easier in a specific case.

Multiply the first equation by 2: 4x - 2y = 6. Add this to the second equation: (4x - 2y = 6) + (3x + 2y = 8) = (7x = 14), so x = 2. Plug this value of x into either of the original equations: 2(2) - y = 3, so y = 1.

Subtract the second equation from the first: (x - 3y = 7) - ( x + 4y = -7) = (0x - 7y = 14). So y = -2. Plug that value of y into either of the original equations: x - 3(-2) = 7, so x + 6 = 7, and x = 1.

Multiply the first equation by 4: 20x + 12y = 48. Multiply the second equation by 5: 20x - 25y = 85. Subtract either new equation from the other: (20x + 12y = 48) - (20x - 25y = 85) = (0x + 37y = -37, so y = -1. Plug that value of y into either of the original equations: 5x + 3(-1) = 12, so x = 3.

From the first equation, b = 10 - 3a. Substituting that b value into the second equation, 2a + 4(10 - 3a) = 0, so that 2a - 12a + 40 = 0, and -10a + 40 = 0. Then -10a = -40, so that a = 4. Substituting that a value into the first original equation, 3(4) + b = 10, so that b = -2.

A very simple example: x + y = 5, and x - y = 1. Adding the first equation to the second equation eliminates y and leaves you with 2x = 6, so that x = 3. Using that value for x in either of the original equations lets you see that y = 2.

The substitution method works only with two equations in two unknowns (or three equations in three unknowns, etc.). Because Equation 1 contains three unknowns, you have two equations in three unknowns, and the substitution method will not work.

Solve the first equation for one variable in terms of the other. Then insert that value into the second equation to solve for the other variable.

Add the equations together, giving you 5x = 20, so that x = 4. Plug that x-value into either original equation: 3(4) + y = 10, or 12 + y = 10, so that y = -2.

Use the second equation to solve for y: y = 30 - 5x. Substitute that y-value in the first equation: 3x + 4(30 - 5x) = 52. Then 3x + 120 - 20x = 52, and 120 - 17x = 52. So 68 = 17x, and x = 4. Plug that x-value into either original equation: 5(4) + y = 30, and y = 30 - 20 = 10.

Using the substitution method, solve for either variable in terms of the other. For example, in the second equation, add 2x to both sides, and y= 2x + 11. Now plug that y-value into the first equation: 3x + 2(2x + 11) = 29. Then 3x + 4x + 22 = 29. So 7x + 22 = 29. Subtract 22 from both sides: 7x = 7, and x =1. Now plug that x-value into either original equation: 3(1) + 2y = 29. So 3 + 2y = 29. Subtract 3 from both sides: 2y = 26, and y = 13. (It would have been slightly easier to use the elimination method to solve this problem, but this article is about the substitution method, so we used that.)

Using the substitution method, in the first equation we can add y to both sides and subtract 10 from both sides, so that y = 2x - 10. Then substituting that y-value into the second equation: 5x + 3(2x - 10) = 10, so 5x + 6x - 30 = 10, or 11x - 30 = 10. 11x = 40, so that x = 40/11. Then plug that x-value into either original equation: 2(40/11) - y = 10. 80/11 - y = 10. Add y to both sides, and subtract 10 from both sides: y = -30/11. Strange values for x and y, but if you use them in either original equation, you find they work!

Use the substitution method. We're given the value of y in terms of x. Plug that y-value into the first equation: 2x + (2x - 6) = 4. 4x - 6 = 4, so 4x = 10, and x = 10/4 = 5/2. Then y = 2(5/2) - 6 = 5 - 6 = -1.

The substitution method is not too useful in this case. Here's a better way: first perform the multiplication to arrive at 3x + 3y = 7y - 7x and 15x - 5y = x + 3. Combine terms in each: 10x = 4y and 14x = 5y + 3. Now solve each of those equations for the same variable. Let's solve for y: the first is y = 10x / 4, and the second is y = (14x - 3) / 5. Set those expressions equal to each other (because they both equal y): 10x / 4 = (14x - 3) / 5. Get rid of the fractions by multiplying both sides by 20: 50x = 56x - 12. Then -6x = -12, and x = 2. Plug that x-value into either of the expressions for y: y = 10(2) / 4 = 5.

The substitution method is not very useful in this case. Instead, first do the multiplication: 3x + 3y = 7y - 7x, and 15x - 5y = x + 3. Now combine terms in each equation: 10x = 4y, and 14x = 5y + 3. Now solve each equation for one of the variables. Let's solve for y: y = 10x / 4 in the first equation, and y = (14x - 3) / 5 in the second equation. Now set those two expressions equal to each other (because they both equal y): 10x / 4 = (14x - 3) / 5. Get rid of the fractions by multiplying both sides of the equation by 20: 50x = 56x - 12. Then -6x = -12, and x = 2. Plug the x-value into either original equation to find that y = 5.

It is. Plug the value of c in. 2(12b - 15) + 3b = -3. 24b - 30 + 3b = -3. 27b - 30 = -3. 27b = 27. b = 1. Plug that back in to the equation of c. 12(1) - 15 = c. 12 - 15 = c. c = -3.

Any simultaneous equation can be solved using elimination. Using this way, you'll be able to add variables together to simplify, then plug these results back into one of the equations to gather the other variable.

If x = 6, y = 9, and a = 10, then this would be 6(9) - 10. 6(9) = 54. 54 - 10 = 44. Your answer is 44.

You will either use substitution method (to solve for x in terms of y) or use the elimination method (to remove the x so that you only have the y).

The elimination method consists of adding or subtracting two equations to cancel out one of the variables, allowing you then to find the value of the other variable. You then plug that value back into either of the original equations to find the value of the variable you initially eliminated. Here’s an example: x + y = 12, and y - x = 2. Add these two equations together: 2y = 14. y = 7. Then x + 7 = 12. x = 12 - 7 = 5.

The elimination method is perfect for this case. (See Solve Simultaneous Equations Using Elimination Method .) First multiply the first equation by 2: 14m - 2n = 10. Now add this new equation to the second original equation: 22m = 22. So m = 1. Plug this m-value into either original equation: 7(1) - n = 5. Then 7 - n = 5, and 2 = n.

Multiply the first equation by -2. You get -6f + 8g = -2. Now add with the second equation. You get 2g = 3, so g = 3/2. Plug that back into any of the equations. 3f - 4(3/2) = 1, 3f - 6 = 1. 3f = 7. f = 7/3.

Using the elimination method is easiest in this case. Multiply the first equation by 7 and the second equation by -3. You get 21x - 28y = -42, and -21x + 6y = -24. Now add those two equations together. -22y = -66. Now divide both sides by -22: y = 3. Plug that back into either of the original equations: 7x - 2(3) = 8. 7x - 6 = 8. 7x = 14. x = 2.

From the first equation we see that x = 11 - 2y. Substitute that x-value into the second equation: 3(11 - 2y) + y = 18. Then 33 - 6y + y = 18, and 33 - 5y = 18. Add 5y to both sides, and subtract 18 from both sides: 15 = 5y. So y = 3. Plug that y-value into either original equation: x + 2(3) = 11, or x + 6 = 11, and x = 5.

From the first equation we see that x = 11 - 2y. Substitute that x-value into the second equation: 3(11 - 2y) + y = 18. Then 33 - 6y + y = 18. So 33 - 5y = 18. Now add 5y to both sides, and subtract 18 from both sides: 15 = 5y, and 3 = y. Plug that y-value into either of the original equations: x + 2(3) = 11, or x + 6 = 11. Then x = 5.

Using the substitution method: solve the first equation for y: add y to both sides, and subtract 2x from both sides: y = 10 - 2x. Now plug that y-value into the second equation: 3x - 2(10 - 2x) = 1. Then 3x - 20 + 4x = 1, and 7x - 20 = 1. Add 20 to both sides: 7x = 21, and x = 3. Plug that x-value into either of the original equations: 2(3) = 10 - y. 6 = 10 - y, and y = 4.

When you have one equation in two variables, all you can do is solve for either variable in terms of the other variable. In this case, y = 5-x, and x = 5-y. (The article above is about solving two simultaneous equations written in the same two variables.)