Nataka nipate majibu kwa kingereza

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11. a) Simplify the following:

11. a) i) $2x - 3y + 4x + 5y$

Group like terms.
$(2x + 4x) + (-3y + 5y)$
Combine like terms.
$6x + 2y$
The simplified expression is:
$\boxed{6x + 2y}$
11. a) ii) $3(x+2) - 2(x-1)$
Distribute the numbers outside the parentheses.
$(3 \times x) + (3 \times 2) - (2 \times x) - (2 \times -1)$ $3x + 6 - 2x + 2$
Group like terms.
$(3x - 2x) + (6 + 2)$
Combine like terms.
$x + 8$
The simplified expression is:
$\boxed{x + 8}$
11. a) iii) $\frac{1}{2} (4x - 6y) + \frac{1}{3} (9x + 3y)$
Distribute the fractions into the parentheses.
$\left(\frac{1}{2} \times 4x\right) - \left(\frac{1}{2} \times 6y\right) + \left(\frac{1}{3} \times 9x\right) + \left(\frac{1}{3} \times 3y\right)$ $2x - 3y + 3x + y$
Group like terms.
$(2x + 3x) + (-3y + y)$
Combine like terms.
$5x - 2y$
The simplified expression is:
$\boxed{5x - 2y}$
11. b) Solve the following equations:

11. b) i) $3x - 5 = 10$
Add 5 to both sides of the equation.
$3x - 5 + 5 = 10 + 5$ $3x = 15$
Divide both sides by 3.
$\frac{3x}{3} = \frac{15}{3}$ $x = 5$
The solution is:
$\boxed{x = 5}$
11. b) ii) $2(x+3) = 14$
Divide both sides by 2.
$\frac{2(x+3)}{2} = \frac{14}{2}$ $x + 3 = 7$
Subtract 3 from both sides.
$x + 3 - 3 = 7 - 3$ $x = 4$
The solution is:
$\boxed{x = 4}$
11. b) iii) $\frac{x}{4} + 2 = 5$
Subtract 2 from both sides of the equation.
$\frac{x}{4} + 2 - 2 = 5 - 2$ $\frac{x}{4} = 3$
Multiply both sides by 4.
$\frac{x}{4} \times 4 = 3 \times 4$ $x = 12$
The solution is:
$\boxed{x = 12}$
11. c) Solve the following simultaneous equations using substitution method:
$x + y = 7 \quad \text{(1)}$
$2x - y = 2 \quad \text{(2)}$
Solve equation (1) for $y$ .
$y = 7 - x \quad \text{(3)}$
Substitute equation (3) into equation (2).
$2x - (7 - x) = 2$ $2x - 7 + x = 2$
Combine like terms and solve for $x$ .
$3x - 7 = 2$ $3x = 2 + 7$ $3x = 9$ $x = \frac{9}{3}$ $x = 3$
Substitute the value of $x$ back into equation (3) to find $y$ .
$y = 7 - 3$ $y = 4$
The solution is:
$\boxed{x = 3, y = 4}$
11. d) Solve the following simultaneous equations using elimination method:
$2x + 3y = 12 \quad \text{(1)}$
$x - y = 1 \quad \text{(2)}$
Multiply equation (2) by 3 to make the coefficients of $y$ opposites.
$3(x - y) = 3(1)$ $3x - 3y = 3 \quad \text{(3)}$
Add equation (1) and equation (3).
$(2x + 3y) + (3x - 3y) = 12 + 3$ $5x = 15$
Solve for $x$ .
$x = \frac{15}{5}$ $x = 3$
Substitute the value of $x$ back into equation (2) to find $y$ .
$3 - y = 1$ $-y = 1 - 3$ $-y = -2$ $y = 2$
The solution is:
$\boxed{x = 3, y = 2}$
11. e) Solve the following quadratic equation by factorization method:
$x^2 + 5x + 6 = 0$
Find two numbers that multiply to 6 and add up to 5.
The numbers are 2 and 3 ( $2 \times 3 = 6$ and $2 + 3 = 5$ ).
Rewrite the middle term using these two numbers.
$x^2 + 2x + 3x + 6 = 0$
Factor by grouping.
$x(x + 2) + 3(x + 2) = 0$
Factor out the common binomial $(x+2)$ .
$(x + 2)(x + 3) = 0$
Set each factor equal to zero and solve for $x$ .
$x + 2 = 0 \quad \text{or} \quad x + 3 = 0$ $x = -2 \quad \text{or} \quad x = -3$
The solutions are:
$\boxed{x = -2 \quad \text{or} \quad x = -3}$

Nataka nipate majibu kwa kingereza

Step-by-step solution

Group like terms.

Combine like terms.

Distribute the numbers outside the parentheses.

Group like terms.

Combine like terms.

Distribute the fractions into the parentheses.

Group like terms.

Combine like terms.

Add 5 to both sides of the equation.

Divide both sides by 3.

Divide both sides by 2.

Subtract 3 from both sides.

Subtract 2 from both sides of the equation.

Multiply both sides by 4.

Solve equation (1) for $y$ .

Substitute equation (3) into equation (2).

Combine like terms and solve for $x$ .

Substitute the value of $x$ back into equation (3) to find $y$ .

Multiply equation (2) by 3 to make the coefficients of $y$ opposites.

Add equation (1) and equation (3).

Solve for $x$ .

Substitute the value of $x$ back into equation (2) to find $y$ .

Find two numbers that multiply to 6 and add up to 5.

Rewrite the middle term using these two numbers.

Factor by grouping.

Factor out the common binomial $(x+2)$ .

Set each factor equal to zero and solve for $x$ .

Still stuck on this one?

More Mathematics Questions

Nataka nipate majibu kwa kingereza

Step-by-step solution

Group like terms.

Combine like terms.

Distribute the numbers outside the parentheses.

Group like terms.

Combine like terms.

Distribute the fractions into the parentheses.

Group like terms.

Combine like terms.

Add 5 to both sides of the equation.

Divide both sides by 3.

Divide both sides by 2.

Subtract 3 from both sides.

Subtract 2 from both sides of the equation.

Multiply both sides by 4.

Solve equation (1) for yyy.

Substitute equation (3) into equation (2).

Combine like terms and solve for xxx.

Substitute the value of xxx back into equation (3) to find yyy.

Multiply equation (2) by 3 to make the coefficients of yyy opposites.

Add equation (1) and equation (3).

Solve for xxx.

Substitute the value of xxx back into equation (2) to find yyy.

Find two numbers that multiply to 6 and add up to 5.

Rewrite the middle term using these two numbers.

Factor by grouping.

Factor out the common binomial (x+2)(x+2)(x+2).

Set each factor equal to zero and solve for xxx.

Still stuck on this one?

More Mathematics Questions

Solve equation (1) for $y$ .

Combine like terms and solve for $x$ .

Substitute the value of $x$ back into equation (3) to find $y$ .

Multiply equation (2) by 3 to make the coefficients of $y$ opposites.

Solve for $x$ .

Substitute the value of $x$ back into equation (2) to find $y$ .

Factor out the common binomial $(x+2)$ .

Set each factor equal to zero and solve for $x$ .