0580 · 2.5Equations

Equality, algebraic language and linear equations

0:00Two cinema tickets can cost the same as three snacks, even though their individual prices are different. An equation records that equality. Solving it reveals which unknown values make the statement true. In this video, we will connect that idea to fractions, simultaneous equations, quadratics and rearranging formulas. You need to recognise like terms, expand brackets and factorise simple expressions. You also need to work confidently with negative numbers and fractions.

0:26We will use those skills without reteaching them completely. Our main focus is why each step preserves the original equality. An expression represents a quantity, while an equation states that two expressions are equal. A formula expresses a relationship between quantities, usually using letters. A solution is a value that makes an equation true. Think of equality as a balanced scale that must remain balanced throughout your working. Adding the same quantity to both sides preserves equality.

0:54Subtracting the same amount does too, as does multiplying or dividing by the same number, provided it is not zero. These operations let us undo what has happened to the unknown. Division by zero is never allowed, so we must watch denominators carefully. Constructing algebra begins by deciding what a letter represents. For two consecutive even numbers, let the first be 2n, where n is an integer. The next is 2n + 2, because consecutive even numbers differ by two.

1:21Their product is 2n multiplied by the whole expression 2n + 2. That product is an expression, not an equation, because no equality has been stated. If their product is eighty, we can equate our expression to eighty. A formula works similarly, but usually describes a general relationship. For example, rectangle perimeter equals twice its length plus twice its width. Solve 3x + 4 = 10 by undoing the addition first.

1:48Subtracting four from both sides gives 3x = 6. Divide both sides by three to obtain x = 2. Substitution checks the result, because three multiplied by two plus four equals ten. This is a linear equation in one unknown because the unknown appears only to the first power. The aim is to isolate that unknown. Saying that a term moves across and changes sign is only shorthand. The real reason is that we perform the same operation on both sides.

2:15Consider 5 − 2x = 3 multiplied by the bracket x + 7. Expanding the right side gives 5 − 2x = 3x + 21. Add 2x to both sides, giving 5 = 5x + 21. Subtract twenty-one to obtain −16 = 5x. Dividing both sides by five gives x = −16/5. The negative answer is perfectly valid unless the context rules it out. When expanding, multiply every term inside the bracket.

2:44Three multiplied by x plus seven is 3x + 21, not 3x + 7.

Solving equations with fractions

2:51Fractions do not change the balancing principle. For x over three plus one over two equals five over six, multiply every term by six. This gives 2x + 3 = 5. Subtracting three gives 2x = 2, and dividing by two gives x = 1. Six works because it is a common multiple of all three numerical denominators. Multiplying the whole equation clears the fractions without changing its solutions. A common mistake is multiplying only one term on a side.

3:20Every term must receive the multiplier, including terms outside fractions. Algebraic denominators require one extra precaution. In x over the expression 2x + 1 = 4, the denominator cannot equal zero. Therefore, x cannot be −1/2. Record this restriction before multiplying both sides by 2x + 1. Clearing the denominator gives x = 4 multiplied by the bracket 2x + 1. Expanding gives x = 8x + 4. Subtract 8x from both sides to obtain −7x = 4.

3:54Division gives x = −4/7. This answer does not equal the excluded value, so it remains possible. Substituting it makes the denominator −1/7. −4/7 ÷ −1/7 = 4. Checking the original equation catches forbidden values and mistakes introduced during manipulation.

Constructing simultaneous equations and elimination

4:14Simultaneous equations describe two conditions that must hold at the same time. One equation with two unknowns often allows many possible pairs. A second equation can identify the pair satisfying both conditions. Constructing these equations means translating each separate piece of information into an equality. Suppose two adult tickets and three child tickets cost nineteen pounds. Three adult tickets and two child tickets cost twenty-one pounds. Let a represent the adult price and c the child price, both in pounds.

4:43Our equations are 2a + 3c = 19, and 3a + 2c = 21. Elimination removes one unknown by making its coefficients match. Multiply the first equation by three to obtain 6a + 9c = 57. Multiplying the second by two gives 6a + 4c = 42. Subtract the second new equation from the first, leaving 5c = 15. Dividing by five gives c = 3. Substitute into the first original equation to obtain 2a + 9 = 19.

5:15Subtract nine, giving 2a = 10, then divide by two to find a equals five. Checking the second equation gives 15 + 6 = 21. The prices are five pounds for an adult and three pounds for a child. Both equations must be satisfied by the same pair. Substitution is another method, where you express one unknown using the other and replace it. We will use that approach when one equation is non-linear.

Factorising quadratics and nonlinear systems

5:41A quadratic equation contains a squared unknown, with no higher power after simplification. Writing it with zero on one side makes several solving methods possible. A quadratic can have two distinct real solutions, one repeated real solution, or no real solutions. We will begin with factorisation, where the reason for setting zero is especially clear. To solve x² − 5x + 6 = 0, factorise the left side. We need two numbers whose product is six and whose sum is negative five.

6:10Negative two and negative three work, giving the brackets x − 2 and x − 3. Their product equals zero. A product is zero only if at least one factor is zero. Therefore, x − 2 = 0, or x − 3 = 0. Solving these gives x = 2 or x = 3. This reasoning would not work if the product equalled six instead of zero. Fractional equations can lead to quadratics when we clear their denominators.

6:36Consider two over the quantity x + 2, plus three over the quantity 2x − 1, equals one. Here, x cannot be negative two or 1/2. Multiply every term by the product of both denominators. The left side becomes two multiplied by two x minus one, plus three multiplied by x plus two. The right side becomes x plus two multiplied by two x minus one. Expanding gives 4x − 2 + 3x + 6 = 2x² + 3x − 2.

7:06Collecting the left terms gives 7x + 4. Subtract 7x and four to obtain 0 = 2x² − 4x − 6. Divide by two, giving x² − 2x − 3 = 0. Factorisation gives x minus three multiplied by x plus one equals zero. Therefore, x = 3 or negative one. Neither answer is excluded by the original denominators. At three, the fractions become two fifths and three fifths, which add to one.

7:35At negative one, they become two and negative one, again totalling one. A fractional equation can have more than one valid solution. Now solve y = 5 − x simultaneously with x² + y² = 13. Replace y in the second equation with 5 − x. This gives x² plus the square of 5 − x = 13. Expanding produces x² + 25 − 10x + x² = 13. Collecting terms and subtracting thirteen gives 2x² − 10x + 12 = 0.

8:08Divide by two to obtain x² − 5x + 6 = 0. Factorising gives x minus two multiplied by x minus three equals zero. Hence x = 2 or three. When x = 2, the linear equation gives y = 3. For x equal to three, it gives y equal to two. Keep these values paired, because swapping or combining them carelessly could produce a false solution. Each pair satisfies both equations, and the squared values add to thirteen.

Completing squares and quadratic formula

8:37Not every quadratic factorises neatly using integers. Completing the square rewrites a quadratic using 1² bracket and a constant. The key pattern is that x + p, all squared, expands to x squared plus two p times x plus p squared. Half the coefficient of x therefore tells us the number inside the bracket. For x² + 6x + 2, half of six is three. Squaring x + 3 gives x² + 6x + 9.

9:05Subtract nine and add the original two to preserve the expression. Its completed square form is therefore the square of x + 3, minus seven. To solve the corresponding equation, set that completed square expression equal to zero. Adding seven gives the square of x + 3 = 7. Taking square roots gives x + 3 equals positive or −√7. Thus, x = −3 plus or minus √7. Both signs matter because a positive number and its negative have the same square.

9:35These answers are exact surd forms, rather than rounded decimals. If the squared term equalled zero, there would be one repeated solution. If it equalled a negative number, there would be no real solution. When the coefficient of x² is not one, factor it out of the quadratic and linear terms first. 2x² + 8x + 1 becomes twice the bracket x² + 4x, plus one. Inside, replace the bracket with the square of x + 2, minus four.

10:03Simplifying gives twice the square of x + 2, minus seven. The quadratic formula packages completing the square into a general result. Start with a times x squared plus b times x plus c equals zero. Dividing by a and completing the square produces the squared bracket shown. Solving that square gives negative b plus or minus the square root of b squared minus four a times c, all over two a.

10:27For 2x² − 3x − 1 = 0, a is two, b is negative three, and c is negative one. The quantity under the root is 9 − −8, giving seventeen. Substitution gives x = 3 plus or minus √17, all over four. These are the exact solutions. If decimals are requested, the answers are approximately one point seven eight and negative zero point two eight one, to three significant figures.

10:55Keep full calculator accuracy until the final answer. The entire numerator must be divided by four. Also, negative b becomes positive three because b itself is negative. The quantity under the square root is called the discriminant. A positive value gives two distinct real solutions. Zero gives one repeated real solution, while a negative value gives no real solutions. This follows directly from the square root in the formula. Fractional equations and the formula often meet in one question.

11:24Solve x over the quantity x + 2, equals three over the quantity x − 6. Here, x cannot be negative two or six. Multiplying both sides by both denominators gives x times the bracket x − 6, equals three times the bracket x + 2. Expanding gives x² − 6x on the left, and 3x + 6 on the right. Rearranging gives x² − 9x − 6 = 0, which does not factorise.

11:52In the formula, a is one, b is negative nine and c is negative six. The discriminant is 81 + 24, which is one hundred and five. So x = 9 plus or minus √105, all over two. Neither value is negative two or six, so both are valid. To three significant figures, they are nine point six two and negative zero point six two three. On Paper two, leave them in this exact surd form.

Rearranging formulas and exam techniques

12:18Changing the subject of a formula uses the same balancing operations as solving an equation. The subject is the letter isolated on one side. Instead of finding one numerical answer, we express that letter using the others. If it appears twice, collect its terms and factorise before dividing. Make x the subject of y = 3x + 2, all over x − 1. Multiply by x − 1 to obtain y multiplied by x minus one equals three x plus two.

12:45Expanding gives y times x minus y equals three x plus two. Collecting the x terms gives y times x minus three x equals y plus two. Factor out x to obtain x multiplied by y minus three equals y plus two. Divide by y − 3, giving x = y + 2, all over y − 3. This requires y not to equal three. Dividing too early would leave x on both sides rather than isolate it.

13:11A power or root requires its inverse operation. From A equals π × r², divide by pi to isolate r². Taking the square root gives r equals the square root of A over pi. Because r represents a radius, we choose the positive root rather than a negative length. If y equals the square root of two x plus three, square both sides first. This gives y² = 2x + 3.

13:35Subtract three, then divide by two, obtaining x = y² − 3, all over two. The original square root also tells us that y cannot be negative. In examinations, commands include construct, solve, write in completed square form, and change the subject. Show equivalent equations line by line so your reasoning is visible. Paper two is non-calculator, so balancing, elimination, factorisation and exact surd work must be manageable without a calculator.

14:04Paper four allows a calculator, but clear algebra still matters. The quadratic formula is provided in the List of formulas, but you must identify its coefficients correctly. Give exact answers when surd form is requested. Otherwise, follow the stated accuracy, usually three significant figures when rounding is needed. Never round intermediate calculations, and check candidate solutions against original denominators. The central idea is preserving equality while making the unknown easier to isolate.

14:32Fractions require denominator restrictions, and simultaneous equations require values that satisfy both conditions together. Quadratics connect factorisation, completed squares and the quadratic formula. Changing the subject applies the same reasoning to a general relationship rather than one numerical problem.