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

"First term" and "constant difference" do not apply to simultaneous equations. (They are used in arithmetic sequences.)

Mathematics knows no national boundaries. There is more than one method of solving simultaneous equations, and all methods are known and "useful" in every country. Apparently you were not taught the elimination method in school.

The process is the same as outlined above, except that you would have to clear a fraction when solving for a variable. For example, you may see something like 5x/2 = 10, in which case you would just multiply both sides of that equation by 2 and then divide both sides by 5.

The article above shows one method. wikiHow has several other articles on the subject. Find them by entering "solve simultaneous equations" in the search bar above.

Add the equations together to cancel the y's: 5x = 27. Divide both sides by 5 to get x = 27/5. Plug that x-value into either original equation to find that y = 21/5. Express your answers as mixed numbers if you prefer.

In addition to the article above, see Solve Simultaneous Equations Using Substitution Method and Solve Simultaneous Equations Graphically .

We don't care whether there is addition or subtraction (or both) within a particular equation. All we want to do is add one equation to the other or subtract one equation from the other. The object in either case is to eliminate one variable so we can solve for the other variable.

Multiply the second equation by 2: 10x + 2y = 74. Add that new equation to the first original equation: 17x = 119, so x = 7. Plug that x-value into either original equation: 5(7) + y = 37. So 35 + y = 37, and y = 2.

If you have simultaneous equations like 2x - y = -1 and 3x - 2y = -3, where the signs in front of the variables in one equation are the same as the corresponding signs in the other equation, the solution in this case is to subtract one equation from the other (rather than adding them together). First perform multiplication in order to make the coefficients of one of the variables the same in each equation (unless they're already the same). If we multiply the first equation by 2, it becomes 4x - 2y = -2. Now we can subtract either equation from the other to eliminate the y-terms, and it becomes easy to solve for x. Then plug the x-value into either original equation to find the y-value. (If subtracting eliminates the x-terms, you would first solve for y and then for x.)

See Solve Simultaneous Equations Using Substitution Method .

Multiply the first equation by 4: 8y - 4x = 12. Now add that new equation to the second equation: 11y = 22. So y = 2. Now plug that y-value into either original equation: 2(2) - x = 3. Then 4 - x = 3, and x = 1.

Let's rearrange the second equation as 3y + 4x = 10. Then multiply by 3 on the first equation, and -2 on the second equation, You get 6y -3x = 9, and -6y + -8x = -20. Now add the two equations together. The 6y and -6y cancel out. -11x = -11. Divide by -11 on both sides. x = 1. Now plug that back into either of the original equations. 2y - 1 = 3. Add 1 on both sides. 2y = 4. Divide by 2 on both sides: y = 2.

There’s not really a correct way or wrong way, but substitution is a little easier when there are variables with the coefficients of 1. If there are no variables with coefficients of 1, probably the elimination method would be best.

Yes. It doesn't matter if you use substitution or elimination method, both methods should give you the same answer.

To solve simultaneous equations with fractions, if you want to get rid of them, multiply the equations by the product of the two denominators.

Since the n terms are identical, just subtract either equation from the other to eliminate the n terms: (3m - 4n = 1) - (m - 4n = 11). The result is 2m = -10, and m = -5. Now plug that m-value into either of the original equations: (3)(-5) - 4n = 1. So -15 - 4n = 1. Add 15 to both sides: -4n = 16, and n = -4.

Using the elimination method, multiply the first equation by 2, then subtract that new equation from the second equation, thus eliminating the y values, and you get x = -8. Plug that x-value into either of the original equations, and you find that y = 26.

Using the elimination method, multiply the first equation by 2, and then subtract that new equation from the second equation. That eliminates the y-terms, and we find that x = -8. Plug that x-value into either of the original equations to see that y = 26.

The elimination method is perfect for this. Simply add the two equations together to eliminate the y terms: 10x = -5, and x = -½. Plug that x-value into either original equation to find that y = 1½.

Multiply the first equation by 3 and the second equation by -2. You get 6x + 15y = 3, and -6x + 4y = -6. Now add. 19y = -3. Divide by 19 on both sides: Y = -3/19. Plug that back in to any of the equations. 3x - 2(-3/19) = 3. 3x + 6/19 = 3. Subtract 6/19 on both sides. 3x = 51/19. Multiply by 1/3: X = 51/57.

Yes, it's very possible and simple.

We know that, exterior angle + interior adjacent angle = 180°. So, if the polygon has n sides, then. Sum of all exterior angles + Sum of all interior angles = n × 180°. So, sum of all exterior angles = n × 180° - Sum of all interior angles. Sum of all exterior angles = n × 180° - (n -2) × 180°.

3x+y=14(1) X+2y=3(2) ÷ 3x+y+x+2y=14+3 3x=17 divide both side by 3 ÷3 ÷3 X=5 the values of x is 5 3x+y=14 (1) 3(5)+y=14(1) 15+y=14(1) +y=14-15 +Y=14 Y=14 3(5)+14=14 15-14=14(1).

See Solve Simultaneous Equations Using Substitution Method .

There are two ways to do this: (1) Multiply the first equation by 2, and then subtract either equation from the other. The result of this is b = 7. Plug that b value into either of the original equations to find that a = -2. (2) Multiply the second equation by 2, and then subtract either equation from the other. The result of this is a = -2. Plug this a value into the either of the original equations to find that b = 7. Either method gives the same answer.

See Solve Simultaneous Equations Using Substitution Method .

Multiply the first equation by 2, then subtract either equation from the other in order to eliminate the x terms. Then solve for y. Plug the y value into either original equation in order to solve for x. (Instead, you could solve for x first by eliminating the y terms. Do this by multiplying the first equation by 3 and the second equation by 5. Then subtract either new equation from the other.) (You would add the equations together instead of subtracting one from the other if one x term was positive and the other negative, or if one y term was positive and the other negative.)

That pair of equations is not solvable. The second equation is just the first equation multiplied by 3, and that situation does not lead to a solution. (There is a trivial (useless) solution, which is that m = 0.)

Add these two equations together in order to eliminate the h-terms: 3g = -3. g = -1. Now plug that g-value into either original equation to solve for h: 2(-1) + h = 2. -2 + h = 2. h = 4.

Multiply the first equation by 6 and the second equation by -7. You get 42r - 18x = -18, and -42r - 63x = -42. Now add. -81x = -60. Divide by 81 on both sides: X = 81/60 (which reduces to 27/20). Plug that back in to any of the equations. 7r - 3(27/20) = -3. 7r - 81/20 = -3. Add 81/20 to both sides 7r = -21/20. Multiply by 1/7 on both sides. R = -21/140.