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.

Use the elimination method. Subtract the first equation from the second. -y = -3, so y = 3. Plug that y-value into either of the original equations: x + 2(3) = 10. x + 6 = 10. x = 10 - 6 = 4.

You are given the value of y in terms of x in the first equation. Plug that value into the second equation to solve for x. Then use the value of x in either equation to solve for y.

Add 2y to both sides of the first equation, so that x = 1 + 2y. Use that x-value in the second equation: 2(2y + 1) - 3y = 4. Then 4y + 2 - 3y = 4. So y + 2 = 4, and y = 2. Now plug that y-value into either of the original equations: for example, x - 2(2) = 1, or x - 4 = 1, and x = 5.

You can work with fractional terms if you want. If not, you can multiply the equation by an appropriate number to eliminate the fraction(s).

This would be easier by elimination. However, by substitution, you solve either equation for one variable in terms of the other variable. Then use that value by substituting it into the other equation. For instance, using the first equation, we find that x = (11 - 2y) / 3. Now plug this x-value into the second original equation: -2[(11 - 2y) / 3] + 7y = 26. Then [(-22 + 4y) / 3] + 7y = 26, and (-22 + 4y)/ 3 + 21y / 3 = 26. So (-22 + 4y +21y) / 3 = 26. Then -22 + 25y = (3)(26) = 78, and 25y = 100, so y = 4. Plug that y-value into either of the original equations to get x = 1.

You can solve simultaneous equations three ways: graphing, substitution, or elimination. Graph by putting both the equations in slope-intercept form and see where the lines intersect. Substitute by solving for one variable in terms of the other. Eliminate by adding or subtracting two equations to eliminate one variable and solve for the other.

Since this article is about the substitution method, we'll use that. In the second equation, we see that y = 15 - 4x. Plug that y-value into the first equation: 2x + 3(15 - 4x) = 11. Then 2x + 45 - 12x = 11, and 45 - 10x = 11. Add 10x to both sides, and subtract 11 from both sides: 34 = 10x. Divide both sides by 10: x = 3.4. Plug that x-value into either original equation: 4(3.4) + y = 15. So 13.6 + y = 15. Then y = 15 - 13.6 = 1.4. (It would have been a little easier to use the elimination method with this problem, but the results would have been the same.)

Using the substitution method, use either equation to solve for one of the variables in terms of the other. Here it's easiest to solve the second equation for y: y = -6x - 4. Now plug that y-value into the first equation: 7x - 2(-6x - 4) + 11 = 0. Then 7x + 12x + 8 + 11 = 0. Combine terms: 19x + 19 = 0. So 19x = -19, and x = -1. Now plug that x-value into either of the original equations: 7(-1) - 2y + 11 = 0. Then -7 - 2y + 11 = 0. So -2y + 4 = 0, and -2y = -4, and y = 2.

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.

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.

Yes, all linear equations (such as these) are solvable by substitution. However, as is often the case, it's easier to solve by the elimination method. (In this case, it's much easier!) First multiply the first equation by 3 and the second equation by 2. Then subtract either new equation from the other new equation. That will eliminate the x-terms and make y easily solvable. Then plug the y value into either original equation to solve for x. (Alternatively, you could multiply the first original equation by 7 and the second by 5 and then subtract either equation from the other to eliminate the y-terms and make x easily solvable. Then plug the x-value into either original equation to solve for y.)

Use the elimination method: subtract the first equation from the second in order to eliminate the x terms and solve for y. Then -4y = -4, and y = 1. Plug that y-value into either original equation: x - 1 = 2, so x = 3.

Solve the first equation for y, and then plug the y-value into the second equation to get the x-value. In this instance, however, that process is unnecessarily unwieldy. It would be much easier to use the elimination method: first multiply the first equation by 4 to get 24x + 4y = 80. Then subtract that equation from the original second equation to get rid of the y terms: -13x = -63, so x = 63/13. Plug that x-value into either original equation to get the value of y: 8(63/13) + y = 20, and 504/13 + y = 20. y = 20 - 504/13 = -244/13. Awkward numbers, but the elimination process is easier than the substitution method in this case.

This is an awkward system of equations, but because the two x terms are identical, you can use the elimination method to make it a little easier. Subtract either equation from the other to eliminate the x terms. For example, subtracting the second equation from the first gives us 11y = -19, and y = -19/11. Plug that value into either original equation. For example, 7x +5(-19/11) = 10. So 7x - 95/11 = 10. Then 7x = 10 + 95/11 = 110/11 + 95/11 = 205/11. Divide both sides by 7 to get x = 205/77. Plug that x value into either of the original equations: for example, 7(205/77) - 6y = (205/11) - 6y = 29. So (205/11) - 29 = 205/11 - 319/11 = -114/11 = 6y, and y = -114/66 = -57/33. Awkward numbers, but if you plug the x and y values into either original equation, they do work.

Add the two equations together: (x - y) + (x + y) = 4 + 6. Then 2x = 10, which means x = 5. Plug that value into either original equation: 5 - y = 4. So y = 1. (Adding two equations together is always a legitimate thing to do and doesn't change the value of any unknown.)

Plug the x value (5) into the first equation. (That's the substitution.) So 5 + 3y = -1. Subtract 5 from both sides of this new equation: 3y = -1 - 5 = -6. Divide both sides by 3: y = -2. So x = 5, and y = -2.

It would be a little easier in this case to use the elimination method, but let's use the substitution method: Since both expressions equal the same thing (0), set them equal to each other: 9p + 3q + 2 = 5p + 4q + 2. Subtract 2 from both sides: 9p + 3q = 5p + 4q. Subtract 5p from both sides, and subtract 3q from both sides: 4p = q. Now substitute that q value into either of the original equations. Let's do the first: 9p + 3(4p) + 2 = 0. Then 9p + 12p + 2 = 0. Combine terms and subtract 2 from both sides: 21p = -2. Then p = -2/21, and q = 4(-2/21) = -8/21. Check by plugging those p and q values into either original equation.