Respuesta :
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
