Question
Solve the simultaneous equations.
3x + 5y = 24
x + 7y = 56
▶️Answer/Explanation
To solve the simultaneous equations:
\(3x + 5y = 24\)
\(x + 7y = 56\)
We’ll use the method of substitution. Let’s solve equation (2) for \(x\):
From equation (2), we have:
\(x = 56 – 7y\)
Now, substitute this value of \(x\) into equation (1):
\(3x + 5y = 24\)
\(3(56 – 7y) + 5y = 24\)
\(168 – 21y + 5y = 24\)
\(-16y = 24 – 168\)
\(-16y = -144\)
\(y = \frac{-144}{-16}\)
\(y = 9\)
Now that we have \(y\), substitute it back into the equation for \(x\):
\(x = 56 – 7y\)
\(x = 56 – 7 \cdot 9\)
\(x = 56 – 63\)
\(x = -7\)
So, the solution to the simultaneous equations is:
\(x = -7\)
\(y = 9\)
Question
Solve the simultaneous equations.
5x + 2y = 16
3x – 4y = 7
▶️Answer/Explanation
To solve the simultaneous equations:
\(5x + 2y = 16\)
\(3x – 4y = 7\)
We’ll use the method of elimination. Let’s manipulate both equations to cancel out one of the variables. First, let’s multiply equation (2) by 2 so that when we add it to equation (1), the coefficients of \(y\) will cancel each other:
\(5x + 2y = 16\)
\(2 \cdot (3x – 4y) = 2 \cdot 7\)
\(5x + 2y = 16\)
\(6x – 8y = 14\)
Now, add equation (1) and equation (2):
\((5x + 2y) + (6x – 8y) = 16 + 14\)
\(11x – 6y = 30\)
Now, solve this equation for one of the variables, say \(x\):
\(11x = 6y + 30\)
\(x = \frac{6y + 30}{11}\)
\(5x + 2y = 16\)
\(5 \cdot \frac{6y + 30}{11} + 2y = 16\)
\(\frac{30y + 150}{11} + 2y = 16\)
\(\frac{30y + 150 + 22y}{11} = 16\)
\(\frac{52y + 150}{11} = 16\)
\(52y + 150 = 176\)
\(52y = 26\)
\(y = \frac{26}{52}\)
\(y = \frac{1}{2}\)
Now that we have \(y\), substitute it back into the equation for \(x\):
\(x = \frac{6y + 30}{11}\]
\(x = \frac{6 \cdot \frac{1}{2} + 30}{11}\]
\(x = \frac{3 + 30}{11}\]
\(x = \frac{33}{11}\]
\(x = 3\)
So, the solution to the simultaneous equations is:
\(x = 3\)
\(y = \frac{1}{2}\)
Question
(a) s = 4t + 3u
Calculate s when t = 2.6 and u = –0.4 .
▶️Answer/Explanation
To calculate the value of \(s\) when \(t = 2.6\) and \(u = -0.4\) in the equation \(s = 4t + 3u\), you simply substitute the given values into the equation and perform the calculations.
\(t = 2.6\) and \(u = -0.4\)
\(s = 4 \cdot 2.6 + 3 \cdot (-0.4)\)
\(s = 10.4 – 1.2\)
\(s = 9.2\)
Therefore, when \(t = 2.6\) and \(u = -0.4\), the value of \(s\) is \(9.2\).
(b) Solve 5x – 7 = 10.
▶️Answer/Explanation
To solve the equation \(5x – 7 = 10\), follow these steps:
\(5x – 7 + 7 = 10 + 7\)
\(5x = 17\)
\(\frac{5x}{5} = \frac{17}{5}\)
\(x = \frac{17}{5}\)
So, the solution to the equation is \(x = \frac{17}{5}\), which can also be expressed as a decimal:
\(x \approx 3.4\)
Question
(a) Solve the simultaneous equations.
You must show all your working.
4x + 2y = 31
6x – 2y = 34
(b) Factorise 14p2+ 21pq.
▶️Answer/Explanation
(a) To solve the simultaneous equations:
\(4x + 2y = 31\)
\(6x – 2y = 34\)
We’ll use the method of elimination. Let’s add equation (1) and equation (2) to cancel out the term with \(y\):
\(4x + 2y = 31\)
\(6x – 2y = 34\)
\((4x + 2y) + (6x – 2y) = 31 + 34\)
\(10x = 65\)
\(x = \frac{65}{10}\)
\(x = 6.5\)
Now that we have \(x\), substitute it back into equation (1) to solve for \(y\):
\(4x + 2y = 31\)
\(4 \cdot 6.5 + 2y = 31\)
\(26 + 2y = 31\)
\(2y = 5\)
\(y = \frac{5}{2}\)
\(y = 2.5\)
So, the solution to the simultaneous equations is:
\(x = 6.5\)
\(y = 2.5\)
(b) To factorise \(14p^2 + 21pq\), you can factor out the common factor, which is \(7p\):
\(7p(2p + 3q)\)
So, the factorised form of \(14p^2 + 21pq\) is \(7p(2p + 3q)\).