We read your photo as:

A) Simplify: (y2+4y+3)/(2y2+7y+5) b) Express as a single fraction: (2k-2t)/15 ÷ (k-t)/3 c) Solve: (x+4)/2 - (2x+1)/3

Mathematics

ScanSolve AI Answer

Answers

6 parts
a

y+32y+5\frac{y+3}{2y+5}

b

25\frac{2}{5}

c

x=10x=10

d

x=1,y=−2x=1, y=-2

e(i)

(x+1)(x+7)

e(ii)

-(4b-3)(b+2)

(3-4b)(b+2)

Step-by-step solution

21 steps

Here are the solutions to the questions.

QUESTION FOUR

Part a· 3 steps

Simplify: y2+4y+32y2+7y+5\frac{y^2+4y+3}{2y^2+7y+5}

  1. Factorize the numerator.

    To factor y2+4y+3y^2+4y+3, find two numbers that multiply to 3 and add to 4. These numbers are 1 and 3.

    y2+4y+3=(y+1)(y+3)y^2+4y+3 = (y+1)(y+3)

  2. Factorize the denominator.

    To factor 2y2+7y+52y^2+7y+5, find two numbers that multiply to 2×5=102 \times 5 = 10 and add to 7. These numbers are 2 and 5.

    2y2+7y+5=2y2+2y+5y+52y^2+7y+5 = 2y^2+2y+5y+5 =2y(y+1)+5(y+1)= 2y(y+1)+5(y+1) =(2y+5)(y+1)= (2y+5)(y+1)

  3. Simplify the fraction by canceling common factors.

    (y+1)(y+3)(2y+5)(y+1)=y+32y+5\frac{(y+1)(y+3)}{(2y+5)(y+1)} = \frac{y+3}{2y+5}

    The simplified expression is y+32y+5\boxed{\frac{y+3}{2y+5}}.

Part b

25\frac{2}{5}

Express as a single fraction: 2k−2t15÷k−t3\frac{2k-2t}{15} \div \frac{k-t}{3}

  1. Rewrite the division as multiplication by the reciprocal of the second fraction.

    2k−2t15×3k−t\frac{2k-2t}{15} \times \frac{3}{k-t}

  2. Factorize the numerator of the first fraction.

    2(k−t)15×3k−t\frac{2(k-t)}{15} \times \frac{3}{k-t}

  3. Multiply the fractions and simplify by canceling common factors.

    2(k−t)×315×(k−t)=6(k−t)15(k−t)\frac{2(k-t) \times 3}{15 \times (k-t)} = \frac{6(k-t)}{15(k-t)}

    Assuming k≠tk \neq t, we can cancel (k−t)(k-t).

    =615= \frac{6}{15}

  4. Reduce the fraction to its simplest form.

    615=2×35×3=25\frac{6}{15} = \frac{2 \times 3}{5 \times 3} = \frac{2}{5}

    The expression as a single fraction is 25\boxed{\frac{2}{5}}.

Part c

x=10x=10

Solve: x+42−2x+13=0\frac{x+4}{2} - \frac{2x+1}{3} = 0

  1. Find a common denominator for the fractions, which is 6.

  2. Multiply each term by the common denominator to eliminate the fractions.

    6(x+42)−6(2x+13)=6(0)6 \left(\frac{x+4}{2}\right) - 6 \left(\frac{2x+1}{3}\right) = 6(0) 3(x+4)−2(2x+1)=03(x+4) - 2(2x+1) = 0

  3. Expand and simplify the equation.

    3x+12−4x−2=03x+12 - 4x-2 = 0 −x+10=0-x+10 = 0

  4. Solve for xx.

    −x=−10-x = -10 x=10x = 10

    The solution is x=10\boxed{x=10}.

Part d

x=1,y=−2x=1, y=-2

Solve the simultaneous equations: 2x+3y=−42x+3y=-4 and x−y=3x-y=3

  1. Label the equations.

    (1) 2x+3y=−42x+3y=-4
    (2) x−y=3x-y=3

  2. From equation (2), express xx in terms of yy.

    x=y+3x = y+3

  3. Substitute this expression for xx into equation (1).

    2(y+3)+3y=−42(y+3)+3y=-4 2y+6+3y=−42y+6+3y=-4 5y+6=−45y+6=-4 5y=−4−65y = -4-6 5y=−105y = -10 y=−105y = \frac{-10}{5} y=−2y = -2

  4. Substitute the value of yy back into the expression for xx from Step 2.

    x=−2+3x = -2+3 x=1x = 1

    The solution is x=1,y=−2\boxed{x=1, y=-2}.

Part e(i)(x+1)(x+7)

x+8x+7x+8x+7

Factorise the following expressions;

Assumption: The expression is intended to be x2+8x+7x^2+8x+7, as x+8x+7x+8x+7 simplifies to 9x+79x+7 which is a linear expression and cannot be factorised into binomials.

  1. For x2+8x+7x^2+8x+7, find two numbers that multiply to 7 and add to 8.

    These numbers are 1 and 7.

  2. Write the factored form.

    x2+8x+7=(x+1)(x+7)x^2+8x+7 = (x+1)(x+7)

    The factorised expression is (x+1)(x+7)\boxed{(x+1)(x+7)}.

Part e(ii)-(4b-3)(b+2)

6−5b−4b26 - 5b - 4b^2

  1. Rearrange the expression in standard quadratic form (ab2+bb+cab^2+bb+c).

    −4b2−5b+6-4b^2 - 5b + 6

  2. Factor out −1-1 to make the leading coefficient positive.

    −(4b2+5b−6)-(4b^2 + 5b - 6)

  3. Factor the quadratic expression 4b2+5b−64b^2 + 5b - 6.

    Find two numbers that multiply to 4×(−6)=−244 \times (-6) = -24 and add to 5. These numbers are 8 and −3-3.

    4b2+8b−3b−64b^2 + 8b - 3b - 6 4b(b+2)−3(b+2)4b(b+2) - 3(b+2) (4b−3)(b+2)(4b-3)(b+2)

  4. Combine with the factored out −1-1.

    −(4b−3)(b+2)-(4b-3)(b+2)

    This can also be written as (3−4b)(b+2)(3-4b)(b+2).
    The factorised expression is −(4b−3)(b+2)\boxed{-(4b-3)(b+2)} or (3−4b)(b+2)\boxed{(3-4b)(b+2)}.

Was this helpful?

Try one yourself

A new problem that tests the same skill.

Still stuck on this one?

Ask a follow-up, and the answer starts from this question.

Got a different question?Ask your own question

More Mathematics Questions

QSimplify: a) (y2+4y+3)/(2y2+7y+5) b) (2k−2t)/15 ÷ (k-t)/3 c) (x+4)/2 − (2x+1)/3QSimplify the following: 1. (3a+4b)/(10) + (5a-b)/(4); 2. (6y−5x)/(10) − (2x+y)/(5) + x/(3); 3…QSimplify the following algebraic fractions: 1) 3/5x2 - 3/2x, 2) 1/y2 - 3/2y − 1/3y, 3) 3/4t2 - 5/6t2…QA. Simplify 3 − 2/(x−3) + 1/(x−3)2 b. Simplify x/(3y-x) − y/(3x) + 1QSimplify the following algebraic expressions: 1) (6a2 b2)/(3a5 b3) 2) (45×3 b7)/(3×4 b2) 3) (21t3 y…
Browse by topic:AlgebraPre-Algebra
See all Mathematics questions →
Got a syllabus? Photograph it and I'll tell you what's due and what's worth the most.
Handwritten step-by-step solution preview

Handwritten Step-by-Step Solution

Get this as a handwritten image on WhatsApp — like a tutor wrote it for you.

Try on WhatsApp