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:

Your answers are below ↓

Step-by-step 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:
[tex]\left \{ {{4x-y=9} \atop {xy=-2}} \right.[/tex]

Rearrange like terms to the same side of the equation:

[tex]\left \{ {{-y=9-4x} \atop {xy=-2}} \right.[/tex]

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

[tex]\left \{ {{y=-9+4x} \atop {xy=-2}} \right.[/tex]

Substitute the unknown quantity into the elimination:

[tex]x(-9+4x)=-2[/tex]

Apply Multiplication Distribution Law:

[tex]-9x+4x^2=-2[/tex]

Reorder the equation:

[tex]4x^2-9x=-2[/tex]

Divide the equation by the coefficient of the quadratic term:

[tex]\frac{1}{4}(4x^2)+\frac{1}{4}(-9x)=\frac{1}{4}*(-2)\\[/tex]

Calculate:

[tex]x^2-\frac{9x}{4}=-\frac{1}{2}[/tex]

Add one term in order to complete the square:

[tex]x^2-\frac{9x}{4}+(\frac{9}{4}*\frac{1}{2})^2=-\frac{1}{2}+(\frac{9}{4}*\frac{1}{2})^2[/tex]

Calculate:

[tex]x^2-\frac{9x}{4}+(\frac{9}{8} )^2=-\frac{1}{2} +(\frac{9}{8} )^2[/tex]

Factor the expression using [tex]a^2$\pm$2ab+b^2=(a$\pm$b)^2[/tex]:

[tex](x-\frac{9}{8} )^2=-\frac{1}{2} +(\frac{9}{8} )^2[/tex]

Simplify using exponent rule with the same exponent rule: [tex](ab)^n=a^n*b^n[/tex]

[tex](x-\frac{9}{8} )^2=-\frac{1}{2} +\frac{9^2}{8^2}[/tex]

Calculate the power:

[tex](x-\frac{9}{8} )^2=-\frac{1}{2}+\frac{81}{64}[/tex]

Find common denominator and write the numerators above the denominator:

[tex](x-\frac{9}{8} )^2=\frac{-32+81}{64}[/tex]

Calculate the first two terms:

[tex](x-\frac{9}{8} )^2=\frac{49}{64}[/tex]

Rewrite as a system of equations:

[tex]x-\frac{9}{8} =\sqrt{\frac{49}{64} }[/tex] or [tex]x-\frac{9}{8} =-\sqrt{\frac{49}{64} }[/tex]

Rearrange unknown terms to the left side of the equation:

[tex]x=\sqrt{\frac{49}{64} } +\frac{9}{8}[/tex]

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

[tex]x=\frac{\sqrt{49} }{\sqrt{64} } +\frac{9}{8}[/tex]

Factor and rewrite the radicand in exponential form:

[tex]x=\frac{\sqrt{7^2} }{\sqrt{8^2} } +\frac{9}{8}[/tex]

Simplify the radical expression:

[tex]x=\frac{7}{8} +\frac{9}{8}[/tex]

Write the numerators over the common denominator:

[tex]x=\frac{7+9}{8}[/tex]

Calculate the first two terms:
[tex]x=\frac{16}{8}[/tex]

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

[tex]x=2[/tex]

Rearrange unknown terms to the left side of the equation:

[tex]x=-\sqrt{\frac{49}{64} } +\frac{9}{8}[/tex]

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

[tex]x=-\frac{\sqrt{49} }{\sqrt{64} }+\frac{9}{8}[/tex]

Factor and rewrite the radicand in exponential form:
[tex]x=-\frac{\sqrt{7^2} }{\sqrt{8^2} } +\frac{9}{8}[/tex]

Simplify the radical expression:

[tex]x=-\frac{7}{8} +\frac{9}{8}[/tex]

Write the numerators over common denominator:

[tex]x=\frac{-7+9}{8}[/tex]

Calculate the first two terms:

[tex]x=\frac{2}{8}[/tex]

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

[tex]x=\frac{1}{4}[/tex]

Find the union of solutions:

[tex]x=2[/tex] or [tex]x=\frac{1}{4}[/tex]

Substitute the unknown quantity into the elimination:

[tex]y=-9+4*2[/tex]

Calculate the first two terms:

[tex]y=-9+8[/tex]

Calculate the first two terms:

[tex]y=-1[/tex]

Substitute the unknown quantity into the elimination:

[tex]y=-9+4*\frac{1}{4 }[/tex]

Reduce the expression to the lowest term:

[tex]y=-9+1[/tex]

Calculate the first two terms:

[tex]y=-8[/tex]

Write the solution set of equations:
[tex]\left \{ {{x=2} \atop {y=-1}} \right.[/tex] or [tex]\left \{ {{x=\frac{1}{4} } \atop {y=-8}} \right.[/tex]   -------> 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              

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