Question

1: solve the following pair of equations simultaneously using the method stated.

a) 2x-3y = 5 and 3x+4y = 6 (elimination method)

b) 4x-y = 9 and 3xy = -6 (substitution method)

c) y=x^2 - 2x and y = 2x -3 (substitution method)​

Answer 1

Your answers are below ↓

Explanation:

Given ↓

A) 2x-3y = 5 and 3x+4y = 6 ( The method this has to be solved in is the elimination method. )

Now using these,

(1)×3 - (2)×2 = 6x + 9y - 6x - 8y = 15 - 12

therefore,

y = 3

putting the value of y in eqn. (1)

2x + 6 = 5

therefore,

x = -1/2

B) y=x^2 - 2x and y = 2x -3 ( The method this has to be solved in is the substitution method. )

Reduce the greatest common factor on both sides of the equation:

`\left \{ {{4x-y=9} \atop {xy=-2}} \right.`

Rearrange like terms to the same side of the equation:

`\left \{ {{-y=9-4x} \atop {xy=-2}} \right.`

Divide both sides of the equation by the coefficient of the variable:

`\left \{ {{y=-9+4x} \atop {xy=-2}} \right.`

Substitute the unknown quantity into the elimination:

`x(-9+4x)=-2`

Apply Multiplication Distribution Law:

`-9x+4x^2=-2`

Reorder the equation:

`4x^2-9x=-2`

Divide the equation by the coefficient of the quadratic term:

`(1)/(4)(4x^2)+(1)/(4)(-9x)=(1)/(4)*(-2)\\`

Calculate:

`x^2-(9x)/(4)=-(1)/(2)`

Add one term in order to complete the square:

`x^2-(9x)/(4)+((9)/(4)*(1)/(2))^2=-(1)/(2)+((9)/(4)*(1)/(2))^2`

Calculate:

`x^2-(9x)/(4)+((9)/(8) )^2=-(1)/(2) +((9)/(8) )^2`

Factor the expression using
`a^2$\pm$2ab+b^2=(a$\pm$b)^2`:

`(x-(9)/(8) )^2=-(1)/(2) +((9)/(8) )^2`

Simplify using exponent rule with the same exponent rule:
`(ab)^n=a^n*b^n`

`(x-(9)/(8) )^2=-(1)/(2) +(9^2)/(8^2)`

Calculate the power:

`(x-(9)/(8) )^2=-(1)/(2)+(81)/(64)`

Find common denominator and write the numerators above the denominator:

`(x-(9)/(8) )^2=(-32+81)/(64)`

Calculate the first two terms:

`(x-(9)/(8) )^2=(49)/(64)`

Rewrite as a system of equations:

`x-(9)/(8) =\sqrt{(49)/(64) }` or
`x-(9)/(8) =-\sqrt{(49)/(64) }`

Rearrange unknown terms to the left side of the equation:

`x=\sqrt{(49)/(64) } +(9)/(8)`

Rewrite the expression using
`\sqrt[n]{ab} =\sqrt[n]{a} *\sqrt[n]{b}`:

`x=(√(49) )/(√(64) ) +(9)/(8)`

Factor and rewrite the radicand in exponential form:

`x=(√(7^2) )/(√(8^2) ) +(9)/(8)`

Simplify the radical expression:

`x=(7)/(8) +(9)/(8)`

Write the numerators over the common denominator:

`x=(7+9)/(8)`

Calculate the first two terms:

`x=(16)/(8)`

Reduce fraction to the lowest term by canceling the greatest common factor:

`x=2`

Rearrange unknown terms to the left side of the equation:

`x=-\sqrt{(49)/(64) } +(9)/(8)`

Rewrite the expression using
`\sqrt[n]{a} =\sqrt[n]{a} *\sqrt[n]{b}`:

`x=-(√(49) )/(√(64) )+(9)/(8)`

Factor and rewrite the radicand in exponential form:

`x=-(√(7^2) )/(√(8^2) ) +(9)/(8)`

Simplify the radical expression:

`x=-(7)/(8) +(9)/(8)`

Write the numerators over common denominator:

`x=(-7+9)/(8)`

Calculate the first two terms:

`x=(2)/(8)`

Reduce fraction to the lowest term by canceling the greatest common factor:

`x=(1)/(4)`

Find the union of solutions:

`x=2` or
`x=(1)/(4)`

Substitute the unknown quantity into the elimination:

`y=-9+4*2`

Calculate the first two terms:

`y=-9+8`

Calculate the first two terms:

`y=-1`

Substitute the unknown quantity into the elimination:

`y=-9+4*(1)/(4 )`

Reduce the expression to the lowest term:

`y=-9+1`

Calculate the first two terms:

`y=-8`

Write the solution set of equations:

`\left \{ {{x=2} \atop {y=-1}} \right.` or
`\left \{ {{x=(1)/(4) } \atop {y=-8}} \right.` -------> Answer

C) y=x^2 - 2x and y = 2x -3 ( This method this has to be solved in is the substitution method. )

Step 1: We start off by Isolating y in y = 2x - 3

y=2x-3 ----------> ( Simplify )

y+(-y)=2x-3+(-y) ---- > ( Add (-y)on both sides)

0=-3+2x-y

y/1 = 2x-3/1 --------> (Divide through by 1)

y = 2x - 3

We substitute the resulting values of y = 2x - 3 and y = x^2 - 2x

(2 * x - 3) = x^2 - 2x ⇒ 2x -3 = x^2 - 2x ----> ↓

(Substituting 2x - 3 for y in y = x^2 -2x )

Next: Solve (2x - 3 = x^2 - 2x) for x using the quadratic formular method

2x - 3 = x^2 - 2x

x = -b±b^2-4ac/2a Step 1: We use the quadratic formula with ↓

a = -1,b=4,c= - 3

x = -4±(4)^2-4(-1)(-3)/2(-1) Step 2: Substitute the values into the Quadratic Formular

x = -4± 4/ - 2 x = 1 or x = 3 Step 3: Simplify the Expression & Separate Roots

x = 1 or x = 3 ------- ANSWER

Substitute 1 in for x in y = 2x - 3 then solve for y

y = 2x - 3

y = 2 · (1) - 3 (Substituting)

y = -1 (Simplify)

Substitute 3 for in y = 2x - 3 then solve for y

y = 2x - 3

y = 2 · (3) - 3 (Substituting)

y = 3 (Simplify)

Therefore, the final solutions for y = x^2 -2x; y = 2x - 3 are

x₁ = 1, y₁ = -1

x₂ = 3, y₂ = 3