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2 years ago

16x+14y=8 -63x-14y=133

Mathematics

1 answer:

artcher [175]2 years ago

4 0

Answer:

(-3, 4)

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Equality Properties

Algebra I

Solving systems of equations using substitution/elimination
Solving systems of equations by graphing

Step-by-step explanation:

Step 1: Define Systems

16x + 14y = 8

-63x - 14y = 133

Step 2: Solve for x

Elimination

Combine 2 equations: -47x = 141
Divide -47 on both sides: x = -3

Step 3: Solve for y

Define original equation: 16x + 14y = 8
Substitute in x: 16(-3) + 14y = 8
Evaluate multiplication: -48 + 14y = 8
Add 48 on both sides: 14y = 56
Divide 14 on both sides: y = 4

Step 4: Graph Systems

Check the solution set.

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2 years ago

What two rational expressions sum to 2x+3/x^2-5x+4

Anni [7]

Answer:

$\frac{2x + 3}{(x- 1)(x - 4)} = \frac{-5}{3(x- 1)} + \frac{11}{3(x - 4)}$

Step-by-step explanation:

Given the rational expression: $\frac{2x + 3}{x^2 - 5x + 4}$ , to express this in simplified form, we would need to apply the concept of partial fraction.

Step 1: factorise the denominator

$x^2 - 5x + 4$

$x^2 - 4x - x + 4$

$(x^2 - 4x) - (x + 4)$

$x(x - 4) - 1(x - 4)$

$(x- 1)(x - 4)$

Thus, we now have: $\frac{2x + 3}{(x- 1)(x - 4)}$

Step 2: Apply the concept of Partial Fraction

Let,

$\frac{2x + 3}{(x- 1)(x - 4)}$ = $\frac{A}{x- 1} + \frac{B}{x - 4}$

Multiply both sides by (x - 1)(x - 4)

$\frac{2x + 3}{(x- 1)(x - 4)} * (x - 1)(x - 4)$ = $(\frac{A}{x- 1} + \frac{B}{x - 4}) * (x - 1)(x - 4)$

$2x + 3 = A(x - 4) + B(x - 1)$

Step 3:

Substituting x = 4 in $2x + 3 = A(x - 4) + B(x - 1)$

$2(4) + 3 = A(4 - 4) + B(4 - 1)$

$8 + 3 = A(0) + B(3)$

$11 = 3B$

$\frac{11}{3} = B$

$B = \frac{11}{3}$

Substituting x = 1 in $2x + 3 = A(x - 4) + B(x - 1)$

$2(1) + 3 = A(1 - 4) + B(1 - 1)$

$2 + 3 = A(-3) + B(0)$

$5 = -3A$

$\frac{5}{-3} = \frac{-3A}{-3}$

$A = -\frac{5}{3}$

Step 4: Plug in the values of A and B into the original equation in step 2

$\frac{2x + 3}{(x- 1)(x - 4)} = \frac{A}{x- 1} + \frac{B}{x - 4}$

$\frac{2x + 3}{(x- 1)(x - 4)} = \frac{-5}{3(x- 1)} + \frac{11}{3(x - 4)}$

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3 years ago

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Aloiza [94]

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3 years ago

Read 2 more answers