Photo AI

15 (a) Solve - OCR - GCSE Maths - Question 15 - 2020 - Paper 1

Question icon

Question 15

15-(a)-Solve-OCR-GCSE Maths-Question 15-2020-Paper 1.png

15 (a) Solve. \( \frac{x + 5}{2} = 15 \) (b) Factorise. \( 5a^{2} - 10a \) (c) Solve by factorising. \( x^{2} + 15x + 56 = 0 \)

Worked Solution & Example Answer:15 (a) Solve - OCR - GCSE Maths - Question 15 - 2020 - Paper 1

Step 1

Solve. \( \frac{x + 5}{2} = 15 \)

96%

114 rated

Answer

To solve for ( x ), first multiply both sides by 2:

x+52⋅2=15⋅2⇒x+5=30\begin{align*} \frac{x + 5}{2} \cdot 2 & = 15 \cdot 2 \\ \Rightarrow x + 5 & = 30 \end{align*}

Next, subtract 5 from both sides:

x=30−5=25\begin{align*} x & = 30 - 5 \\ & = 25 \end{align*}

Thus, ( x = 25 ).

Step 2

Factorise. \( 5a^{2} - 10a \)

99%

104 rated

Answer

To factorise the expression ( 5a^{2} - 10a ), identify the common factor:

  • The common factor is ( 5a ).

Now factor out ( 5a ):

5a(a−2)5a(a - 2)

Therefore, the factorised form is ( 5a(a - 2) ).

Step 3

Solve by factorising. \( x^{2} + 15x + 56 = 0 \)

96%

101 rated

Answer

To solve this quadratic equation by factorising, first look for two numbers that multiply to 56 and add up to 15. These numbers are 7 and 8.

Now, we can express the equation as:

(x+7)(x+8)=0(x + 7)(x + 8) = 0

Setting each factor equal to zero gives:

x+7=0⇒x=−7\begin{align*} x + 7 & = 0 \\ \Rightarrow x & = -7 \end{align*}

and

x+8=0⇒x=−8\begin{align*} x + 8 & = 0 \\ \Rightarrow x & = -8 \end{align*}

Thus, the solutions are ( x = -7 ) and ( x = -8 ).

Join the GCSE students using SimpleStudy...

97% of Students

Report Improved Results

98% of Students

Recommend to friends

100,000+

Students Supported

1 Million+

Questions answered

;