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

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How do I find the first term and the constant difference of simultaneous equations?

Donagan

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

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Is this useful in all the countries? Where I live, we work our equations very differently.

Donagan

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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.

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How do I work with fractions in solving simultaneous equations using elimination method?

Donagan

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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.

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If 7x - 2y = 45, and 5x + y = 37, what are x and y?

I_l1ke_gam3s

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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.

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What if both equations are addition?

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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.

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How do you solve simultaneous equations?

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

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How do you solve simultaneous equations?

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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.

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2g + h = 2, and g - h = -5. How do I solve this?

I_l1ke_gam3s

Community Answer

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.

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What if the signs are same?

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Then you subtract one equation from the other instead of adding them together. That will eliminate one variable and allow you to solve for the other variable. For example, x + 3y = 9 and x - y = 7. Subtract the second equation from the first. That gives us 4y = 2, and y = ½. Then we use that y-value to solve for x.

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, solve using elimination method 2x + 3y = 3 and 6x + 10y =24

I_l1ke_gam3s

Community Answer

Multiply the first equation by -3. You get -6x -9y = -9. Then add that to the second equation. y = 15. Plug that back into either of the equations: 2x + 3(15) = 3. 2x + 45 = 3. 2x = -42. x = -21.

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How to solve 3p +4m=8 and 5p-6m=10

I_l1ke_gam3s

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Let’s eliminate the m: multiply the first equation by 3, and the second equation by 2. You get 9p + 12m = 24, and 10p - 12m = 20. Add these two equations together. 19p = 44. p = 44/19. Plug that p-value into either of the original equations: 3(44/19) + 4m = 8. 132/19 + 4m = 8. 4m = 8 - 132/19 = 152/19 - 132/19 = 20/19 = m. Divide both sides by 4: 5/19 = m.

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Is there a possibility that I might solve wrong when I use the substitution method?

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Yes, of course, if you make a mistake. Both the elimination and substitution methods work well if you avoid mistakes.

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Solve these equations: 3x-y=12 and 2x-y=13.

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Subtract the second equation from the first to eliminate the y terms: (3x - y) - (2x - y) = 12 - 13. So 3x - 2x -y + y = -1. That means x = -1. Plug that value of x back into either original equation: (3)(-1) - y = 12. So -3 - y = 12. Add y to both sides, and subtract 12 from both sides: -15 = y. So x = -1, and y = -15. (Plug each of those values into either of the original equations to check your work.)

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The denominator of a fraction is 4 more than the numerator. if both numerator and denominator are increased by 1, the fraction will become 1/2. Find the original fraction.

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Let x be the original numerator. Then (x + 4) is the original denominator. So the original fraction can be designated as x / (x + 4). Now add 1 to both the numerator and the denominator and set that new fraction equal to 1/2: (x + 1) / (x + 4 + 1) = (x + 1) / (x + 5) = 1/2. Now cross-multiply: (2)(x + 1) =(1)(x + 5). So (2x + 2) = (x + 5). Subtract x from both sides, and subtract 2 from both sides: x = 3. So the original numerator is 3, the original denominator is 3 + 4 = 7, and the original fraction is 3/7. (When you add 1 to both the numerator and the denominator, you get 4/8 = 1/2.)

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7r - 3x = -3 , 6r + 9x = 6

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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.

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Solve the simultaneous equations using the elimination method: 2x+5y=1, 3x-2y=3.

I_l1ke_gam3s

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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.

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2m¹-m²=0 and 6m¹-3m²=0

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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.)

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2×-5y=-6 and 4×-3y=-12

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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.)

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What about the substitution method?

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See Solve Simultaneous Equations Using Substitution Method .

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Using the elimination method, solve a+2b=12, and 2a+b=3.

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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.

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Solve using substitution method 4x-6y=16 ,6x+8y=-10

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See Solve Simultaneous Equations Using Substitution Method .

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So my question was 3x+y=14, x+2y=3?

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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).

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How do I find the sum of the exterior angles of a polygon?

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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°.

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Can I solve simultaneous equations using the substitution method?

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Yes, it's very possible and simple.

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