a+2b=13
2a-3b=5
2a+2b = 26
2a-3b = 5
now subtract, and the a goes away, and you can find b.
oops
2a+4b=26
Eq1: a + 2b = 13.
Eq2: 2a - 3b = 5.
Multiply both sides of Eq1 by -2 and the Eqs:
-2a - 4b = -26.
2a - 3b = 5.
Sum: -7b = -21,
b = 3.
In Eq1, replace b with 3 and solve for a.
To solve this system of equations, we can use the method of substitution or elimination. Let's use the method of substitution.
Step 1: Solve one equation for a variable.
Let's solve the first equation for 'a':
a + 2b = 13
a = 13 - 2b
Step 2: Substitute the expression found in step 1 into the other equation.
Substitute '13 - 2b' for 'a' in the second equation:
2(13 - 2b) - 3b = 5
Step 3: Simplify and solve for 'b'.
Distribute the 2 to the terms inside the parentheses:
26 - 4b - 3b = 5
Combine like terms:
26 - 7b = 5
Subtract 26 from both sides:
-7b = 5 - 26
-7b = -21
Divide both sides by -7:
b = -21 / -7
b = 3
Step 4: Substitute the value found in step 3 into one of the original equations and solve for 'a'.
Let's substitute 'b = 3' into the first equation:
a + 2(3) = 13
a + 6 = 13
Subtract 6 from both sides:
a = 13 - 6
a = 7
Therefore, the solution to the system of equations is 'a = 7' and 'b = 3'.