Rewrite the equation in the slope-intercept form y = mx + c.

Question

Rewrite the equation in the slope-intercept form y = mx + c.

Asked on March 26, 2026Mathematics

This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.

Accepted Answer · 2026-03-26T23:16:29.216Z

*1) Given the equation of a line: $\frac{x}{4} + \frac{y}{3} - \frac{1}{2} = 0$. a) Determine the gradient of the line. Step 1: Rewrite the equation in the slope-intercept form $y = mx + c$. $$ \frac{y}{3} = -\frac{x}{4} + \frac{1}{2} $$ Step 2: Multiply by 3 to isolate $y$. $$ y = 3 \left(-\frac{x}{4} + \frac{1}{2}\right) $$ $$ y = -\frac{3}{4}x + \frac{3}{2} $$ Step 3: Identify the gradient $m$. The gradient $m$ is the coefficient of $x$. $$ m = -\frac{3}{4} $$ The gradient of the line is $\boxed{-\frac{3}{4}}$. b) Determine the coordinates of the point where the line cuts the coordinate axes. Step 1: Find the x-intercept (where $y=0$). Substitute $y=0$ into the original equation: $$ \frac{x}{4} + \frac{0}{3} - \frac{1}{2} = 0 $$ $$ \frac{x}{4} - \frac{1}{2} = 0 $$ $$ \frac{x}{4} = \frac{1}{2} $$ $$ x = 4 \times \frac{1}{2} $$ $$ x = 2 $$ The x-intercept is $(2, 0)$. Step 2: Find the y-intercept (where $x=0$). Substitute $x=0$ into the original equation: $$ \frac{0}{4} + \frac{y}{3} - \frac{1}{2} = 0 $$ $$ \frac{y}{3} - \frac{1}{2} = 0 $$ $$ \frac{y}{3} = \frac{1}{2} $$ $$ y = 3 \times \frac{1}{2} $$ $$ y = \frac{3}{2} $$ The y-intercept is $\left(0, \frac{3}{2}\right)$. The coordinates where the line cuts the coordinate axes are $\boxed{(2, 0) \text{ and } \left(0, \frac{3}{2}\right)}$. c) An equation of a line passing through $(a, b)$ and perpendicular to the line above. Step 1: Find the gradient of the given line. From part (a), the gradient of the given line is $m = -\frac{3}{4}$. Step 2: Find the gradient of the perpendicular line. If two lines are perpendicular, the product of their gradients is $-1$. Let $m_{\perp}$ be the gradient of the perpendicular line. $$ m \cdot m_{\perp} = -1 $$ $$ \left(-\frac{3}{4}\right) m_{\perp} = -1 $$ $$ m_{\perp} = \frac{-1}{-\frac{3}{4}} $$ $$ m_{\perp} = \frac{4}{3} $$ Step 3: Use the point-slope form $y - y_1 = m_{\perp}(x - x_1)$ with $(x_1, y_1) = (a, b)$. $$ y - b = \frac{4}{3}(x - a) $$ Step 4: Rearrange into a standard form (optional, but often preferred). $$ 3(y - b) = 4(x - a) $$ $$ 3y - 3b = 4x - 4a $$ $$ 4x - 3y - 4a + 3b = 0 $$ The equation of the perpendicular line is $\boxed{y - b = \frac{4}{3}(x - a) \text{ or } 4x - 3y - 4a + 3b = 0}$. 2) The price tag on an article in a supermarket is 7,000 FCFA. During an auction sale, the price was reduced by 12%. a) At what price will the article be sold? Step 1: Calculate the reduction amount. Reduction = $12\%$ of 7,000 FCFA $$ \text{Reduction} = \frac{12}{100} \times 7000 $$ $$ \text{Reduction} = 12 \times 70 $$ $$ \text{Reduction} = 840 \, \text{FCFA} $$ Step 2: Calculate the selling price. Selling price = Original price - Reduction $$ \text{Selling price} = 7000 - 840 $$ $$ \text{Selling price} = 6160 \, \text{FCFA} $$ The article will be sold for $\boxed{6160 \, \text{FCFA}}$. b) Given that there is a 1% tax on the purchase price, how much will be paid? (Leave your answer to the nearest 5FCFA). Step 1: Calculate the tax amount. The purchase price is the selling price from part (a), which is 6160 FCFA. Tax = $1\%$ of 6160 FCFA $$ \text{Tax} = \frac{1}{100} \times 6160 $$ $$ \text{Tax} = 61.60 \, \text{FCFA} $$ Step 2: Calculate the total amount paid. Total paid = Selling price + Tax $$ \text{Total paid} = 6160 + 61.60 $$ $$ \text{Total paid} = 6221.60 \, \text{FCFA} $$ Step 3: Round the total amount to the nearest 5 FCFA. The nearest multiple of 5 to 6221.60 is 6220. The amount to be paid is $\boxed{6220 \, \text{FCFA}}$. 3) Given that $f(x) = x^3 - 2x^2 - 5x + 6$. a) Show that $(x+2)$ is a factor of $f$. Step 1: Use the Factor Theorem. According to the Factor Theorem, if $(x+2)$ is a factor of $f(x)$, then $f(-2)$ must be equal to 0. Substitute $x = -2$ into $f(x)$: $$ f(-2) = (-2)^3 - 2(-2)^2 - 5(-2) + 6 $$ $$ f(-2) = -8 - 2(4) - (-10) + 6 $$ $$ f(-2) = -8 - 8 + 10 + 6 $$ $$ f(-2) = -16 + 16 $$ $$ f(-2) = 0 $$ Since $f(-2) = 0$, $\boxed{(x+2) \text{ is a factor of } f(x)}$. b) Factorise $f(x)$ completely. Step 1: Divide $f(x)$ by $(x+2)$ to find the quadratic factor. Using synthetic division with root $-2$: ` -2 | 1 -2 -5 6 | -2 8 -6 ------------------ 1 -4 3 0 ` The coefficients of the quotient are $1, -4, 3$. So, the quadratic factor is $x^2 - 4x + 3$. Step 2: Factorise the quadratic expression $x^2 - 4x + 3$. We need two numbers that multiply to 3 and add to -4. These numbers are -1 and -3. $$ x^2 - 4x + 3 = (x - 1)(x - 3) $$ Step 3: Write $f(x)$ as a product of its linear factors. $$ f(x) = (x+2)(x-1)(x-3) $$ The complete factorisation of $f(x)$ is $\boxed{(x+2)(x-1)(x-3)}$.

Accepted Answer · 2026-03-26T23:16:29.216Z

*1) Given the equation of a line: $\frac{x}{4} + \frac{y}{3} - \frac{1}{2} = 0$. a) Determine the gradient of the line. Step 1: Rewrite the equation in the slope-intercept form $y = mx + c$. $$ \frac{y}{3} = -\frac{x}{4} + \frac{1}{2} $$ Step 2: Multiply by 3 to isolate $y$. $$ y = 3 \left(-\frac{x}{4} + \frac{1}{2}\right) $$ $$ y = -\frac{3}{4}x + \frac{3}{2} $$ Step 3: Identify the gradient $m$. The gradient $m$ is the coefficient of $x$. $$ m = -\frac{3}{4} $$ The gradient of the line is $\boxed{-\frac{3}{4}}$. b) Determine the coordinates of the point where the line cuts the coordinate axes. Step 1: Find the x-intercept (where $y=0$). Substitute $y=0$ into the original equation: $$ \frac{x}{4} + \frac{0}{3} - \frac{1}{2} = 0 $$ $$ \frac{x}{4} - \frac{1}{2} = 0 $$ $$ \frac{x}{4} = \frac{1}{2} $$ $$ x = 4 \times \frac{1}{2} $$ $$ x = 2 $$ The x-intercept is $(2, 0)$. Step 2: Find the y-intercept (where $x=0$). Substitute $x=0$ into the original equation: $$ \frac{0}{4} + \frac{y}{3} - \frac{1}{2} = 0 $$ $$ \frac{y}{3} - \frac{1}{2} = 0 $$ $$ \frac{y}{3} = \frac{1}{2} $$ $$ y = 3 \times \frac{1}{2} $$ $$ y = \frac{3}{2} $$ The y-intercept is $\left(0, \frac{3}{2}\right)$. The coordinates where the line cuts the coordinate axes are $\boxed{(2, 0) \text{ and } \left(0, \frac{3}{2}\right)}$. c) An equation of a line passing through $(a, b)$ and perpendicular to the line above. Step 1: Find the gradient of the given line. From part (a), the gradient of the given line is $m = -\frac{3}{4}$. Step 2: Find the gradient of the perpendicular line. If two lines are perpendicular, the product of their gradients is $-1$. Let $m_{\perp}$ be the gradient of the perpendicular line. $$ m \cdot m_{\perp} = -1 $$ $$ \left(-\frac{3}{4}\right) m_{\perp} = -1 $$ $$ m_{\perp} = \frac{-1}{-\frac{3}{4}} $$ $$ m_{\perp} = \frac{4}{3} $$ Step 3: Use the point-slope form $y - y_1 = m_{\perp}(x - x_1)$ with $(x_1, y_1) = (a, b)$. $$ y - b = \frac{4}{3}(x - a) $$ Step 4: Rearrange into a standard form (optional, but often preferred). $$ 3(y - b) = 4(x - a) $$ $$ 3y - 3b = 4x - 4a $$ $$ 4x - 3y - 4a + 3b = 0 $$ The equation of the perpendicular line is $\boxed{y - b = \frac{4}{3}(x - a) \text{ or } 4x - 3y - 4a + 3b = 0}$. 2) The price tag on an article in a supermarket is 7,000 FCFA. During an auction sale, the price was reduced by 12%. a) At what price will the article be sold? Step 1: Calculate the reduction amount. Reduction = $12\%$ of 7,000 FCFA $$ \text{Reduction} = \frac{12}{100} \times 7000 $$ $$ \text{Reduction} = 12 \times 70 $$ $$ \text{Reduction} = 840 \, \text{FCFA} $$ Step 2: Calculate the selling price. Selling price = Original price - Reduction $$ \text{Selling price} = 7000 - 840 $$ $$ \text{Selling price} = 6160 \, \text{FCFA} $$ The article will be sold for $\boxed{6160 \, \text{FCFA}}$. b) Given that there is a 1% tax on the purchase price, how much will be paid? (Leave your answer to the nearest 5FCFA). Step 1: Calculate the tax amount. The purchase price is the selling price from part (a), which is 6160 FCFA. Tax = $1\%$ of 6160 FCFA $$ \text{Tax} = \frac{1}{100} \times 6160 $$ $$ \text{Tax} = 61.60 \, \text{FCFA} $$ Step 2: Calculate the total amount paid. Total paid = Selling price + Tax $$ \text{Total paid} = 6160 + 61.60 $$ $$ \text{Total paid} = 6221.60 \, \text{FCFA} $$ Step 3: Round the total amount to the nearest 5 FCFA. The nearest multiple of 5 to 6221.60 is 6220. The amount to be paid is $\boxed{6220 \, \text{FCFA}}$. 3) Given that $f(x) = x^3 - 2x^2 - 5x + 6$. a) Show that $(x+2)$ is a factor of $f$. Step 1: Use the Factor Theorem. According to the Factor Theorem, if $(x+2)$ is a factor of $f(x)$, then $f(-2)$ must be equal to 0. Substitute $x = -2$ into $f(x)$: $$ f(-2) = (-2)^3 - 2(-2)^2 - 5(-2) + 6 $$ $$ f(-2) = -8 - 2(4) - (-10) + 6 $$ $$ f(-2) = -8 - 8 + 10 + 6 $$ $$ f(-2) = -16 + 16 $$ $$ f(-2) = 0 $$ Since $f(-2) = 0$, $\boxed{(x+2) \text{ is a factor of } f(x)}$. b) Factorise $f(x)$ completely. Step 1: Divide $f(x)$ by $(x+2)$ to find the quadratic factor. Using synthetic division with root $-2$: ` -2 | 1 -2 -5 6 | -2 8 -6 ------------------ 1 -4 3 0 ` The coefficients of the quotient are $1, -4, 3$. So, the quadratic factor is $x^2 - 4x + 3$. Step 2: Factorise the quadratic expression $x^2 - 4x + 3$. We need two numbers that multiply to 3 and add to -4. These numbers are -1 and -3. $$ x^2 - 4x + 3 = (x - 1)(x - 3) $$ Step 3: Write $f(x)$ as a product of its linear factors. $$ f(x) = (x+2)(x-1)(x-3) $$ The complete factorisation of $f(x)$ is $\boxed{(x+2)(x-1)(x-3)}$.