Respuesta :
Answer:
(1, 2)
General Formulas and Concepts:
Pre-Algebra
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
- Left to Right
Equality Properties
- Multiplication Property of Equality
- Division Property of Equality
- Addition Property of Equality
- Subtract Property of Equality
Algebra I
- Solving systems of equations using substitution/elimination
Step-by-step explanation:
Step 1: Define Systems
-4x + 9y = 14
12x - 10y = -8
Step 2: Rewrite Systems
-4x + 9y = 14
- Multiply everything by 3: -12x + 27y = 42
Step 3: Redefine Systems
-12x + 27y = 42
12x - 10y = -8
Step 4: Solve for y
Elimination
- Combine 2 equations: 17y = 34
- Divide 26 on both sides: y = 2
Step 5: Solve for x
- Define equation: 12x - 10y = -8
- Substitute in y: 12x - 10(2) = -8
- Multiply: 12x - 20 = -8
- Isolate x term: 12x = 12
- Isolate x: x = 1
Step-by-step explanation:
HERE,
two equation are,
●-4x+9y=14••••••••••••(equation I)
●12x-10y=-8•••••••••••(equation II)
First multiplying 3 in equation I
we get,
[tex]\bold{3×(-4x+9y=14) }[/tex]
=[tex]\bold{ -12x+27y=42 }[/tex]••(equation III)
Then,
we combine the equationii and equation III.
we get that,
[tex]\bold{12x-10y-12x+27y=-8+42 }[/tex]
[tex]\bold{\cancel{12x}-10y\cancel{-12x}+27y=-8+42 }[/tex]
[tex]\rightsquigarrow[/tex] [tex]\bold{17y=34 }[/tex]
[tex]\rightsquigarrow[/tex] [tex]\bold{ y=\dfrac{34}{17} }[/tex]
[tex]\rightsquigarrow[/tex] [tex]\boxed{ y=2 }[/tex]
Then,
put the value of y in equation II.
WE get,
[tex]\rightsquigarrow[/tex] [tex]\bold{12x-10×2=-8 }[/tex]
[tex]\rightsquigarrow[/tex] [tex]\bold{ 12x-20=-8 }[/tex]
[tex]\rightsquigarrow[/tex] [tex]\bold{ 12x=-8+20 }[/tex]
[tex]\rightsquigarrow[/tex] [tex]\bold{ 12x=12 }[/tex]
[tex]\rightsquigarrow[/tex] [tex]\bold{ x=\dfrac{12}{12} }[/tex]
[tex]\rightsquigarrow[/tex] [tex]\boxed{ x=1 }[/tex]
So,
solution of the two equation (-4x+9y) and (12x-10y=-8) is (1,2)
