Algebra in English · Álgebra en inglés
| English | Español |
|---|---|
| coefficient/ˌkəʊɪˈfɪʃənt/ | coeficiente |
| constant/ˈkɒnstənt/ | constante |
| term/tɜːm/ | término |
| inequality/ɪniːˈkwɒlɪti/ | la desigualdad |
| equation/ɪˈkweɪʒn/ | ecuación |
| identity/aɪˈdentɪti/ | identidad |
| expand/ekˈspænd/ | desarrollar |
| factorise/ˈfæktəraɪz/ | factorizar |
| discriminant/dɪˈskrɪmɪnənt/ | discriminante |
| simultaneous equations/ˌsɪməlˈteɪnɪəs ɪˈkweɪʒnz/ | sistema de ecuaciones simultáneas |
Represent an unknown relationship
- In $5x^2-3x+7$, 5 is the coefficient 系数 of $x^2$, 7 is a constant · constante 常数, and each separated piece is a term 项.
- A variable can represent many possible values, not just a number waiting to be found. Translate the stated relationship before selecting an operation.
In the expression 5x² − 3x + 7, what is 5 called? · En la expresión 5x² − 3x + 7, ¿qué es 5?
A coefficient multiplies a term with a letter in it. The constant here is 7, which has no letter. · Un coeficiente multiplica un término que contiene una letra. La constante aquí es 7, que no tiene letra.
Equality and inequality have different solutions
- An equation 方程 states equality; an identity 恒等式 holds for every allowed value in its domain. An inequality · la desigualdad 不等式 uses a comparison and may have a range of solutions.
- Add or subtract the same amount on both sides. Multiplying or dividing an inequality by a negative number reverses its sign; dividing by a possibly zero expression needs separate care.
(x + 1)² = x² + 2x + 1 is an identity, not an equation to solve. · (x + 1)² = x² + 2x + 1 es una identidad, no una ecuación para resolver.
It is true for every value of x, so there is nothing to solve — it is a fact about the algebra. · Es verdadero para cada valor de x, por lo que no hay nada que resolver — es un hecho sobre el álgebra.
Expand and factorise
- To expand 展开, distribute multiplication across every bracket term. To factorise 因式分解, express a sum as a product.
- A zero product has at least one zero factor. From $(x-4)(x+3)=0$, obtain $x=4$ or $x=-3$ and check both in the original equation.
Expand (x + 3)(x − 5) and give the coefficient of x. · Desarrolla (x + 3)(x − 5) y da el coeficiente de x.
The x terms are −5x and +3x, giving −2x. The full expansion is x² − 2x − 15. · Los términos con x son −5x y +3x, dando −2x. El desarrollo completo es x² − 2x − 15.
Choose a method and check every condition
- For · A favor $ax^2+bx+c=0$ with $a\ne0$, use factorisation or $x=(-b\pm\sqrt{b^2-4ac})/(2a)$. The discriminant 判别式 is $b^2-4ac$: positive, zero and negative give two, one and no distinct real roots respectively.
- Simultaneous equations 联立方程 require a solution satisfying all equations. Elimination or substitution can reveal one solution, no solution or infinitely many; check the original pair.
One solution of x² − 2x − 15 = 0 is negative. What is it? · Una solución de x² − 2x − 15 = 0 es negativa. ¿Cuál es?
Factorising gives (x − 5)(x + 3) = 0, so x = 5 or x = −3. · Factorizando se obtiene (x − 5)(x + 3) = 0, así que x = 5 o x = −3.
Balance two ticket purchases. $3a+2s=46$ and · y $2a+4s=44$. Doubling the first equation gives $6a+4s=92$. Subtraction gives $4a=48$, so $a=12$ and · y $s=5$. Both original totals are satisfied.
Solve −2x > 6. · Resuelve −2x > 6.
Dividing both sides by −2 reverses the sign. Test x = −4: −2(−4) = 8 > 6 ✓. · Dividir ambos lados por −2 invierte el signo. Prueba x = −4: −2(−4) = 8 > 6 ✓.
A quadratic has discriminant b² − 4ac = 0. How many real solutions does it have? · Una cuadrática tiene discriminante b² − 4ac = 0. ¿Cuántas soluciones reales tiene?
The ± in the formula adds and subtracts zero, so both roots are the same value. · El ± en la fórmula suma y resta cero, por lo que ambas raíces tienen el mismo valor.
Interpret roots after solving. A rectangle with width x and length $x+3$ can give roots 5 and negative 8. Only positive dimensions fit the physical problem. Keep both algebraic roots before explaining the selection; practise this on sheet 1.2.
For · A favor $4-2x\leq10$, subtract 4 to get $-2x\leq6$, then divide by negative 2 to get $x\geq-3$. The boundary is included. A single trial value can expose a mistake but does not prove an entire solution set.