## Algebra 2 diagnostic test

**Introduction :-**

In this article we are learn about the algebra 2 diagnostic test concept. Algebra is cluster of math and it process on the algebra 2 diagnostic test. Algebra 2 diagnostic test covers the four basic operation such as addition, subtraction, multiplication and division. The mainly important expression of algebra 2 diagnostic test is variable, constant, coefficient, exponent, word and expression. algebra 2 diagnostic test beside numeral we use character and alphabet in place of unidentified number to make a statement. So, algebra 2 diagnostic test may be regard as an addition of Arithmetic.

**Diagnostic test Chapter for algebra 2:-**

• Primary degree Equations and Inequalities.

• Linear Relations and Functions.

• Systems of Equations and Inequalities.

• Quadratic Functions and Inequalities.

• Polynomial Functions.

• Rational Expressions and Equations.

• Exponential and Logarithmic Relations.

• Sequences and Series.

• Trigonometric Functions.

**Algebra 2 diagnostic test example problems:-**

**Algebra 2 diagnostic test problem 1:-**

Solve –2(4a + 3) + 6(2a – 4) = –42

**Solution:-**

–2(4a + 3) + 6(2a – 4) = –42 Original equation

–8a – 6 + 12a – 24 = –42

4a – 30 = –42

4a = –12

a = –3

The answer is –3.

**Algebra 2 diagnostic test problem 2:-**

Write every equation in standard form. Identify A, B, and C.

6y – 24x + 12 = 0

**Solution:-**

6y – 24x + 12 = 0 Original equation

6y – 24x = –12 Subtract 12 from each side.

4x – y = 2 Divide each side by –6 so that the coefficients have a

GCF of 1 and A ≥ 0.

So, A = 4, B = –1, and C = 2.

**Algebra 2 diagnostic test problem 3:-**

Solve by using substitution and solve the system of equations.

2x – y = –5

3y – 5x = 14

Solve the first equation for y in terms of x.

**Solution:-**

2x – y = –5 First equation

–y = –5 – 2x Subtract 2x from each side.

y = 5 + 2x multiply each side by –1.

Substitute 5 + 2x for y in the second equation and solve for x.

3y – 5x = 14 Second equation

3(5 + 2x) – 5x = 14 Substitute 5 + 2x for y.

15 + 6x – 5x = 14

15 + x = 14 Simplify.

x = –1 Subtract 15 from each side.

Now, substitute the value for x in either original equation and solve for y.

2x – y = -5 First equation

2(–1) – y = -5 Replace x with –1.

–2 – y = -5 Simplify.

–y = -3 Add 2 to each side.

y = 3 multiply each side by –1.

The final solution of the system is (–1, 3).