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Simultaneous equations Simplified Revision Notes

Revision notes with simplified explanations to understand Simultaneous equations quickly and effectively.

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Simultaneous Equations

What are Simultaneous Equations?

  • Simultaneous equations involve two equations with two unknown letters.
  • You solve them together to find the value of the unknown letters.
  • Key Point: The values you find must make both equations balance. This means checking your answers in both equations to ensure correctness.

How to Solve Simultaneous Equations: Step-by-Step

Step 1: Arrange the Equations

Step 2: Align the Equations

Step 3: Choose Your Key Letter

Step 4: Subtract or Add the Equations

Step 5: Solve the Remaining Equation

Step 6: Substitute Back

Step 7: Check Your Answer


Step 1: Arrange the Equations

  • Make sure the equations are in the same format, usually something like:
ax+by=cax+by=c dx+ey=fdx+ey=f

Step 2: Align the Equations

  • Write one equation underneath the other, ensuring the unknown letters line up vertically.

Step 3: Choose Your Key Letter

  • Decide which letter you want to eliminate. This letter will become your Key Letter.
  • Manipulate the equations, if necessary, to make the coefficients of this Key Letter the same in both equations.

Step 4: Subtract or Add the Equations

  • Subtract the equations if the Key Letters have the same sign.
  • Add the equations if the Key Letters have opposite signs.
  • This step should eliminate the Key Letter, leaving you with one equation and one unknown.

Step 5: Solve the Remaining Equation

  • Solve the simplified equation to find the value of the remaining unknown letter.

Step 6: Substitute Back

  • Take the value you found and substitute it back into one of the original equations to solve for the other unknown letter.

Step 7: Check Your Answer

  • Substitute both values into the other original equation to ensure that both equations are satisfied.

infoNote

Example 1: Solving Simultaneous Equations Let's solve the following simultaneous equations:

3x+y=193x+y=19x+y=9x+y=9

Step-by-Step Solution:

  1. Align the Equations:
  • Make sure both equations are written in the same form. Here, both are in the form ax+by=cax+by=c.

  1. Write One Equation Under the Other:
  • Place the second equation directly below the first one:
3x+y=19(Equation1)3x+y=19(Equation 1)x+y=9(Equation2)x+y=9(Equation 2)
  1. Choose the Key Letter:
  • You need to decide which variable (either xx or yy) to eliminate.
  • Notice that the coefficients of yy are already the same (both are +1+1), so we'll choose yy as our Key Letter to eliminate.

  1. Subtract the Equations:
  • Since the coefficients of yy are the same and have the same sign, subtract the second equation from the first:
(3x+y)(x+y)=199(3x+y)−(x+y)=19−9
  • This simplifies to:
2x=102x=10
  1. Solve for xx:
  • Now that y$$ is eliminated, solve for xx:
x=102=5x=\frac{10}{2}=5
  1. Substitute x=5x=5 Back into One of the Original Equations:
  • Use either of the original equations to find the value of y$$. Let's use Equation 22:
5+y=95+y=9
  • Subtract 55 from both sides:
y=4y=4
  1. Final Answer:
  • The solution is x=5x=5 and y=4y=4.

  1. Check the Solution:
  • Substitute x=5x=5 and y=4y=4 back into the original equations to ensure they satisfy both
3(5)+4=15+4=19(Correct for Equation 1)3(5)+4=15+4=19 \quad(Correct\ for\ Equation \ 1)5+4=9(CorrectforEquation2)5+4=9\quad(Correct for Equation 2)

infoNote

Example 2: Solving Simultaneous Equations Let's solve the following simultaneous equations:

3x2y=33x−2y=32x+2y=122x+2y=12

Step-by-Step Solution:

  1. Align the Equations:
  • Make sure both equations are written in the same form, so you can easily compare and manipulate them.

  1. Write One Equation Under the Other:
  • Place the second equation directly below the first one:
3x2y=3(Equation1)3x−2y=3\quad (Equation 1)2x+2y=12(Equation2)2x+2y=12\quad(Equation 2)
  1. Choose the Key Letter:
  • Decide which variable (either xx or yy) to eliminate.
  • Notice that the coefficients of yy are already the same in magnitude (both are 22), but with different signs. This makes it easier to eliminate yy by adding the equations.

  1. Add the Equations:
  • Since the coefficients of yy are the same but with opposite signs, add the two equations to eliminate yy:
(3x2y)+(2x+2y)=3+12(3x−2y)+(2x+2y)=3+12
  • This simplifies to:
5x=155x=15
  1. Solve for xx:
  • Now that yy is eliminated, solve for xx:
x=155=3x=\frac{15}5{=}3
  1. Substitute x=3x=3 Back into One of the Original Equations:
  • Use either of the original equations to find the value of yy. Let's use Equation 22:
2(3)+2y=122(3)+2y=12
  • Simplify and solve for yy:
6+2y=126+2y=122y=126=62y=12−6=6y=62=3y=\frac{6}{2}=3
  1. Final Answer:
  • The solution is x=3x=3 and y=3y=3.

  1. Check the Solution:
  • Substitute x=3x=3 and y=3y=3 back into the original equations to ensure they satisfy both:
3(3)2(3)=96=3(Correct for Equation 1)3(3)−2(3)=9−6=3\quad(Correct\ for\ Equation\ 1)2(3)+2(3)=6+6=12(Correct for Equation 2)2(3)+2(3)=6+6=12\quad(Correct\ for\ Equation\ 2)

infoNote

Example 3: Solving Simultaneous Equations Let's solve the following simultaneous equations:

2x+3y=72x+3y=73x+5y=183x+5y=18

Step-by-Step Solution:

  1. Align the Equations:
  • First, ensure that both equations are written in a comparable form, so you can easily manipulate them.

  1. Write One Equation Under the Other:
  • Place the second equation directly below the first one:
2x+3y=7(Equation 1)2x+3y=7\quad(Equation\ 1)3x+5y=18(Equation 2)3x+5y=18\quad(Equation\ 2)
  1. Adjust the Equations for Elimination:
  • We need to make the coefficients of either xx or yy the same so that we can eliminate one of the variables.
  • Multiply Equation 11 by 33 and Equation 22 by 22 to make the coefficients of xx equal:
(2x+3y)×3=6x+9y=21(Equation 3)(2x+3y)×3=6x+9y=21\quad(Equation \ 3)(3x+5y)×2=6x+10y=36(Equation 4)(3x+5y)×2=6x+10y=36\quad (Equation\ 4)
  1. Eliminate One Variable:
  • Subtract Equation 33 from Equation 44 to eliminate xx:
(6x+10y)(6x+9y)=3621(6x+10y)−(6x+9y)=36−21
  • This simplifies to:
y=15y=15
  1. Solve for the Remaining Variable:
  • Now that we have y=15y=15, substitute this value back into one of the original equations (Equation 11) to find xx:
2x+3(15)=72x+3(15)=7
  • Simplify and solve for xx:
2x+45=72x+45=72x=745=382x=7−45=−38x=382=19x=\frac{−38}{2}=−19
  1. Final Answer:
  • The solution is x=19x=−19 and y=15y=15.

  1. Check the Solution:
  • Substitute x=19x=−19 and y=15y=15 back into the original equations to ensure they satisfy both:
2(19)+3(15)=38+45=7(Correct for Equation 1)2(-19) + 3(15) = -38 + 45 = 7 \quad (\text{Correct for Equation 1}) \\ 3(19)+5(15)=57+75=18(Correct for Equation 2)3(-19) + 5(15) = -57 + 75 = 18 \quad (\text{Correct for Equation 2})

infoNote

Example 4: Solving Simultaneous Equations Given the simultaneous equations:

7x2y=20(Equation 1)7x−2y=−20\quad(Equation\ 1)3x=64y(Equation 2)3x=6−4y\quad(Equation\ 2)

Step-by-Step Solution:

Step 1: Simplify and align the equations

First, let's rewrite the second equation in a form similar to the first:

3x+4y=63x+4y=6

So now we have:

7x2y=20(Equation 1)7x−2y=−20\quad(Equation\ 1)3x+4y=6(Equation 2)3x+4y=6\quad(Equation\ 2)

Step 2: Make the coefficients of one of the variables the same

To eliminate one of the variables, we need the coefficients of either xx or yy to be the same. We can do this by multiplying the first equation by 22:

2×(7x2y)=2×202×(7x−2y)=2×−20

This gives us:

14x4y=40(Equation 3)14x−4y=−40\quad (Equation\ 3)

Now we have:

14x4y=40(Equation 3)14x−4y=−40\quad (Equation\ 3)3x+4y=6(Equation 2)3x+4y=6\quad (Equation \ 2)

Step 3: Add the equations to eliminate yy

Since the coefficients of yy in Equation 22 and Equation 33 are the same but with opposite signs, we can add these two equations to eliminate yy:

(14x4y)+(3x+4y)=40+6(14x−4y)+(3x+4y)=−40+6

Simplifying:

17x=3417x=−34x=2x=−2

Step 4: Substitute x=2xx=−2x into one of the original equations to find yy

Let's use Equation 22 for substitution:

3(2)+4y=63(−2)+4y=6

Simplifying:

6+4y=6−6+4y=64y=124y=12y=3y=3

Final Answer: The solution to the system of equations is:

x=2andy=3x=−2 \quad and\quad y=3


Step 5: Check the solution

Finally, substitute x=2x=−2 and y=3y=3 back into the original equations to ensure they satisfy both:

  • For Equation 11:
7(2)2(3)=20True!7(−2)−2(3)=−20 \quad True!
  • For Equation 22:
3(2)+4(3)=6True!3(−2)+4(3)=6\quad True!

Both equations are satisfied, so the solution is correct!


Simultaneous Equations Involving Quadratics

Simultaneous equations can sometimes involve one equation that is quadratic. This adds an extra step to the process, but don't worry—it's still manageable! Let's look at how to solve these.

Steps to Solve Simultaneous Equations Where One Equation is Quadratic

  1. Re-arrange the linear equation
  2. Substitute
  3. Simplify the resulting quadratic equation
  4. Factorise the quadratic equation
  5. Substitute each solution
  6. Check

  1. Re-arrange the linear equation so that one variable (usually yy or xx) is isolated. This gives you an expression that you can substitute into the quadratic equation.

  1. Substitute this expression into the quadratic equation. This means that wherever you see the isolated variable in the quadratic equation, you replace it with the expression from the linear equation.

  1. Simplify the resulting quadratic equation. You should now have an equation with only one unknown variable. Re-arrange it to make it equal to zero.

  1. Factorise the quadratic equation or use the quadratic formula to solve for the variable. This will give you two possible values.

  1. Substitute each solution back into the original linear equation to find the corresponding value of the other variable.

  1. Check your solutions by substituting both values into the original equations to ensure they satisfy both equations.

infoNote

Worked Example

y=x2(Quadratic Equation)y=x^2\quad (Quadratic\ Equation)y=2x+3 (LinearEquation)y=2x+3\ (Linear Equation)

Step-by-Step Solution:

  1. Isolate yy in the linear equation:
y=2x+3y=2x+3

This is already isolated, so we can use this directly.


  1. Substitute y=2x+3y=2x+3 into the quadratic equation:
2x+3=x22x+3=x^2
  1. Re-arrange the equation to make it equal to zero:
x22x3=0x^2−2x−3=0

This is our quadratic equation to solve.


  1. Factorise the quadratic equation:
(x3)(x+1)=0(x−3)(x+1)=0

So, x=3x=3 or x=1x=−1.


  1. Find yy for each xx value using the linear equation y=2x+3y=2x+3:
  • For x=3x=3:
y=2(3)+3=6+3=9y=2(3)+3=6+3=9
  • For x=1x=−1:
y=2(1)+3=2+3=1y=2(−1)+3=−2+3=1

Final Answers:

[$$ (x,y)=(3,9)\quad or\quad (x,y)=(−1,1)

--- 13. **Check your solutions** by substituting them back into both original equations: - For $(3,9)$: $y=x^2\quad gives\quad 9=3^2=9\quad and\quad y=2x+3\quad gives\quad 9=2(3)+3=9$ Both equations are satisfied. - For $(−1,1)$: $y=x^2\quad gives\quad 1=(−1)^2=1\quad and\quad y=2x+3\quad gives\quad 1=2(−1)+3=1$ ::question{#140044}

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